ohm's law experiment class 10 observation

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Experiment to Verify Ohm's Law

Last updated at April 16, 2024 by Teachoo

Graphical Representation Volt and Amperes

Experiment to Verify Ohm's Law

We take a conductor (Example Nichrome Wire)

We connect it to a circuit containing Voltmeter and Ammeter

When we supply current, we measure reading of Potential Difference with the help of Voltmeter and Electric Current with help of Ammeter

We calculate Ratio of Potential Difference/Electric Current

Now,we increase amount of current,

We again measure reading of Potential Difference and Electric Current and again Calculate Ratio

We note that Ratio Remains the Same

Hence Ohm's Law, which states that Ratio of Potential Difference and Electric Current Remains the same, is verified

Q1. The values of Current (I) flowing through a conductor for the corresponding values of potential difference (V) are given. Plot a graph between V and I.

From the above table

We can see that,

the ratio of 𝑉/𝐼 is always constant.

This gives resistance

The resistance is 25 in above case

Q2. The values of Current (I) flowing through a conductor for the corresponding values of potential difference (V) are given. Plot a graph between V and I.

the ratio of V/I is always  constant.

The resistance is 4 in above case

Q3. The values of Current (I) flowing through a conductor for the corresponding values of potential difference (V) are given. Plot a graph between V and I. Hence. find the resistance.

From the above table We can see that, the ratio of 𝑉/𝐼 is nearly constant. To find resistance, we find the mean of the resistances found. R = (3.2 + 3. 3 + 3.35 + 3.4 + 3.3)/5 R = 16.55/5 R = 3.31 Ω

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Verification of Ohm’s Law experiment with data and graph

In the previous article, we discussed Ohm’s Law of current electricity. In this article, we’re going to perform an experiment for the verification of Ohm’s law. This practical verification of Ohm’s law is very important for the students of grades 10 and 12. This is a lab-based experiment to verify Ohm’s law or Ohm’s law practical.

Aim of the Experiment

Theory of the ohm’s law experiment.

From Ohm’s law , we know that the relation between electric current and potential difference is V = IR

or, \color{Blue}R=\frac{V}{I} ………….. (1)

or, resistivity, \color{Blue}\rho = \frac{RA}{L} ………. (2)

Where A is the cross-section area of the wire. A = πr 2 where r is the radius of the wire. L is the length of the wire.

Apparatus Used

The apparatus used for this experiment –

Circuit Diagram

Here, R is the resistance of the wire, A is the ammeter, V is the Voltmeter, Rh is the rheostat and K is the key. The arrow sign indicates the direction of the current flow in the circuit .

Formula used for the Ohm’s law lab experiment

\color{Blue}R = \frac{V}{I} ………….. (1) and \color{Blue}\rho = \frac{RA}{L} ………. (2)

Experimental data

The least count of Voltmeter = Smallest division of voltmeter = 0.05 Volt

We also need to plot I-V graph to confirm the experimental value of R.

Current versus Voltage graph (Ohm’s Law graph)

If we plot the Current as a function of voltage with the help of the above data then we will get a straight line passing through the origin.

Calculations

Calculation of resistance from the graph.

The inverse of the I-V graph gives the resistance of the wire. Now, from the graph, change in current, ∆I = AB = 0.5 amp corresponding change in voltage, ∆V = BC = 0.5 volt Thus, the Resistance from the graph, R = ∆V/∆I = 0.5/0.5 = 1.00 ohm

Calculation of resistivity of the wire

Length of the wire is, L = 50 cm = 0.5 m Radius of the wire. r = 0.25 mm = 0.25 × 10 -3 m So, the cross-section area of the wire, A = πr 2 = 3.14 × (0.25×10 -3 ) 2 = 0.196 × 10 -6 m 2 Thus from the equation-2 we get the resistivity of the material of the wire is, \rho = (1 × 0.196 ×10 -6 )/0.5 or, \rho = 0.392 × 10 -6 = 3.92 ×10 -7 ohm.m Thus the resistivity of the material of the wire is 3.92 ×10 -7 ohm.m

Final result

The resistance of the wire from the Current-Voltage graph is, R = 1.00 ohm The calculated value of the resistance of the wire is, R = 1.02 ohm. Resistivity of the material of the wire is 3.92 ×10 -7 ohm.m

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CBSE Class 10 Science Lab Manual – Ohm’s Law

January 2, 2023 by Veerendra

Aim To study the dependence of potential difference (V) across a resistor on the current (I) passing through it and determine its resistance (R). Also plot a graph between V and I.

Materials Required A voltmeter and ammeter of suitable range, resistance wire or resistor, some connecting wires, a key, a dry cell (or battery eliminator), a rheostat and a piece of sand paper.

Theory/Principle Ohm’s Law The dependence of current on potential difference can be studied by Ohm’s law. According to Ohm’s law, the potential difference (V) across the ends of a resistor is directly proportional to the current (I) through it provided its temperature remains the same. i.e. V ∝ I or V/I = Constant = R or V = RI Here, R is a constant for the given resistor at a given temperature and is called its resistance. The SI unit of resistance is ohm (Ω). The graph between the potential difference across the two ends of a resistor and the current passing through it is a straight line passing through the origin. The slope of this straight line graph gives the resistance R of the resistor.

• Before starting the experiment, note down the range, least count and zero error (if any) of the ammeter and voltmeter.
• Wipe the ends of the connecting wires with sand paper to remove any insulating layer.

• Connect the key, the rheostat, the ammeter and the resistor in series with the connecting wires to the dry cell, as shown in the circuit diagram [Fig. 1(b)], Ensure that positive (+) terminal of ammeter is joined to the positive (+) terminal of the cell. Make neat and tight connections.
• Connect the voltmeter in parallel to the resistor, as shown in circuit diagram. Ensure that the positive (+) terminal of the voltmeter is joined to the positive (+) terminal of the cell, and the negative (-) terminal of the voltmeter is joined to the negative (-) terminal of the cell.
• Insert the key in the plug to let the current establish in the circuit.
• Adjust the rheostat, so that a small current passes through the resistor, say 0.1 A, so the reading of ammeter will be 0.1 A.
• Read the corresponding value of potential difference from voltmeter.
• Repeat the experiment by adjusting the slider, for the values of current 0.3A, 0.4A, 0.5A etc. and record the corresponding values of potential difference in voltmeter.
• Lastly, plot a graph between V and I taking V along y-axis and I along x-axis.

• From the above calculation and observation, we have concluded that for a resistor, if temperature remains constant then current through the resistor is directly proportional to the potential difference applied across it.
• The value of resistance R of resistor remains the same for all values of current through it. The graph between V and I is a straight line and passes through the origin. This verifies Ohm’s law.

Precautions

• Wires should be thick and their insulation of ends should be removed properly.
• All the connections should be tight otherwise some external resistance may introduce in the circuit.
• Ammeter must be connected in series with the circuit.
• Voltmeter must be connected in parallel with the circuit.
• Zero mark of ammeter and voltmeter should be checked properly.
• There should be tight connections of connecting wires.
• Put the plug in key, only when reading is to be recorded. Take off plug immediately after it, to avoid unnecessary heating of wires.
• Positive terminals of voltmeter and ammeter should be connected to positive terminal of the cell and vice-versa.

Sources Of Error

• There may be a reading error of ammeter and voltmeter.
• Area of cross-section of wire may not be uniform.
• It may be possible that current has flown for longer period of time through the circuit.
• There may be a loose connection in electrical circuit.

Viva – Voce

Question 1. In this experiment, it is advised to take out the key from the plug when the observations are not being taken. Why? [NCERT] Answer: It is to avoid the unnecessary heating of wires. Heating may change the resistance of resistors.

Question 2. Suppose the ammeter (or voltmeter) you are using in this experiment do not have positive (+) and negative (-) terminal markings. How will you use such ammeter (or voltmeter) in the circuit? [NCERT] Answer: First of all we connect ammeter or voltmeter arbitrary in the circuit, if deflection occurs in opposite direction. It means the connection is not proper. Now by interchanging the terminals, we can use it in the circuit.

Question 3. If the resistor of known resistance value is replaced with a nichrome wire of 10 cm length. How do the values of current through the nichrome wire and potential difference across the two ends of it may change? How the values will change, if the replaced wire is of manganin in place of nichrome? [NCERT] Answer: Nichrome wire offers more resistance. So, value of current decreases, while value of potential difference across the wire increases. As manganin offers less resistance than nichrome so, value of current increases and potential difference decreases.

Question 4. Suppose in this experiment, you see that the deflection on ammeter (or voltmeter) scale goes beyond the full scale. What will you infer from such an observation? What will you infer, if the deflection takes place in opposite direction? [NCERT] Answer: If the deflection goes beyond full scale, a higher range of ammeter is required for the circuit. And deflection in opposite direction indicates that terminal has not been joined properly and it should be reversed.

Question 5. Why is it advised to clean the ends of connecting wires before connecting them? [NCERT] Answer: Ends of wire should be cleaned to remove the insulation (or insulating layer) of dust particles since, presence of insulation (which is a non-conducting material) wilt not allow current to flow in the circuit.

Question 6. What do you mean by a battery eliminator? Answer: A device which is used in place of battery or cells, is known as a battery eliminator.

Question 7. What are the necessary conditions for current flow through the conductor? Answer: There are two such conditions:

• There must be a closed circuit.
• There must be a source of electrical energy.

Question 8. What is the composition of nichrome alloy? Answer: The composition of nichrome alloy are as follows: Ni = 68% Cr = 15% Fe = 15.5% Mn = 1.5%

Question 9. Can you explain why there is no current in the circuit, if plug is removed? Answer: As we know air gap gives infinite resistance to the circuit. So, if the plug is removed, there will be no current in the circuit.

Question 10. What do you understand by the term ‘resistance’ of a conductor? Answer: The resistance of a conductor may simply be explained as being the obstruction offered to the passage of electric current by the conductor in the circuit.

Question 11. What do you understand by the term ’emf of a cell’? Answer: The maximum potential difference which exists across the terminals of a cell when no current is drawn out of the cell is called emf (electromotive force) of a cell.

Question 12. What do you understand by the term ‘terminal potential difference of a cell’? Answer: The potential difference which exists across the terminals of a cell when current is being drawn from the cell, is known as terminal potential difference of a cell.

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Electricity of Class 10

The flow of electric current through a conductor depends on the potential difference across its ends. At a particular temperature, the strength of current flowing through it is directly proportional to the potential difference across its ends. This is known as Ohm's Law.

or V ∝ I V = Potential difference

V = RI R = Resistance

or R = V/I, I = Current

Here, R is the constant of proportionality, which depends on size, nature of material and temperature. R is called the electrical resistance or resistance of the conductor.

EXPERIMENTAL VERIFICATION OF OHM'S LAW

• Set up a circuit as shown in figure consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metal.)
• First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them.
• Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.

•  Repeat the above steps using three cells and then four cells in the circuit separately.
• Calculate the ratio of V to I for each pair of potential difference (V) and current (I).
• Plot a graph between V and I, and observe the nature of the graph.

Thus, V/I  is a constant ratio which is called resistance (R). It is known as Ohm’s Law.

RESISTANCE OF A CONDUCTOR:

The electric current is a flow of electrons through a conductor. When the electrons move from one part of the conductor to the other part, they collide with other electrons and with the atoms and ions present in the body of the conductor. Due to these collisions, there is some obstruction or opposition to the flow of electrons through the conductor.

The property of a conductor due to which it opposes the flow of current through it, is called resistance. The resistance of a conductor is numerically equal to the ratio of potential difference across its ends to the current flowing through it.

Resistance = Potential difference/Current, Or R = V/I

UNIT OF RESISTANCE

The S.I. unit of resistance is Ohm (Ω)

1 Ohm (Ω) = 1 volt(1V)/ Ampere(1 A)

The resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is applied across its ends.

CONDUCTORS, RESISTORS AND INSULATORS:

On the basis of their electrical resistance, all the substances can be divided into three groups:

conductors, resistors and insulators.

Conductors:

Those substances which have very low electrical resistance are called conductors. A conductor allows the electricity to flow through it easily. Silver metal is the best conductor of electricity Copper and Aluminium metals are also good conductors. Electric wires are made of Copper or Aluminium because they have very low electrical resistance.

Those substances which have comparatively high electrical resistance, are called resistors. The alloys like nichrome, manganin and constantan (or ureka), all have quite high resistances, so they are used to make those electrical devices where high resistance is required. A resistor reduces the current in the circuit.

Insulators:

Those substances which have infinitely high electrical resistance are called insulators. An insulator does not allow electricity to flow through it. Rubber is an excellent insulator. Electricians wear rubber handgloves while working with electricity because rubber is an insulator and protects them from electric shocks. Wood is also a good insulator.

CAUSE OF RESISTANCE:

There are many free electrons in a conductor. They move randomly when no electric current is passing through it. But when current is passed through it, they being negatively charged, start moving towards positive end of conductor, with a velocity called Drift velocity. During this movement, they collide with atoms, or ions of the conductor and thus their velocity is slowed down. This slow down due to obstruction is called Resistance.

Activity to show that the amount of current through an electric component depends upon its resistance:

•  Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0-5 A range), a plug key and some connecting wires.
•  Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in figure.

•  Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current through the circuit.]
•  Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
• Now repeat the above step with 10 W bulb in the gap XY.
•  You will notice that the ammeter readings differ for different components connected in the gap XY.
• You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

Thus, we come to a conclusion that current through an electric component depends upon its resistance.

FACTORS AFFECTING RESISTANCE OF A CONDUCTOR

Resistance depends upon the following factors:-

(i) Length of the conductor.

(ii) Area of cross-section of the conductor (or thickness of the conductor).

(iii) Nature of the material of the conductor.

(iv) Temperature of the conductor.

Mathematically: It has been found by experiments that:

(i) The resistance of a given conductor is directly proportional to its length i.e.

R ∝ l ….(i)

(ii) The resistance of a given conductor is inversely proportional to its area of cross-section i.e.

R  ∝ 1/A  ….(ii)

From (i) and (ii), R  ∝ 1/A

R = ρ x 1/A…(iii)

Where ρ (rho) is a constant known as resistivity of the material of the conductor. Resistivity is also known as specific resistance.

DEPENDANCY OF RESISTANCE ON TEMPERATURE:

If R 0 is the resistance of the conductor at 0°C and R1 is the resistance of the conductor at t°C then the relation between R0 and R1 is given by,

R 1 = R o (1 + θαΔt) [Here Δt = t – 0 = t]

Here, a = Coefficient of Resistivity, t = temperature in °C

Experiment to show that resistance of a conductor depends on its length, cross section area and nature of its material.

• Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length [marked (1)] and a plug key, as shown in figure.
• Now, plug the key. Note the current in the ammeter.

•  Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2 [marked (2) in the figure].
•  Note the ammeter reading.
•  Now replace the wire by a thicker nichrome wire, of the same length  [marked(3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
•  Instead of taking a nichrome wire, connect a copper wire [marked (4) in figure] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked(1)]. Note the value of the current.
•  Notice the difference in the current in all cases.
•  We notice that the current depends on the length of the conductor.
•  We also observed that the current depends on the area of cross-section of the wire used.

RESISTIVITY:

Resistivity,  ρ = R x A/1….(iv)

By using this formula, we will now obtain the definition of resistivity. Let us take a conductor having a unit area of cross-section of 1 m 2 and a unit length of 1 m. So, putting A= 1 and l = 1 in equation (iv),

Resistivity, ρ = R

The resistivity of a substance is numerically equal to the resistance of a rod of that substance which is 1 metre long and 1 metre square in cross-section.

Unit of resistivity,

The S.I. unit of resistivity is ohm-metre which is written in symbols as Ω - m.

Resistivity of a substance does not depend on its length or thickness. It depends only on the nature of the substance. The resistivity of a substance is its characteristic property. So, we can use the resistivity values to compare the resistances of two or more substances.

 Importance of resistivity:

A good conductor of electricity should have a low resistivity and a poor conductor of electricity should have a high resistivity. The resistivities of alloys are much more higher than those of the pure metals. It is due to their high resistivities that manganin and constantan alloys are used to make resistance wires used in electronic appliances to reduce the current in an electrical circuit.

Nichrome alloy is used for making the heating elements of electrical appliances like electric irons, room-heaters, water-heaters and toasters etc. because it has very high resistivity and it does not undergo oxidation (or burn) even when red-hot.

Effect of temperature on resistivity:

The resistivity of conductors (like metals) is very low. The resistivity of most of the metals increases with temperature. On the other hand, the resistivity of insulators like ebonite, glass and diamond is very high and does not changes with temperature. The resistivity of semi-conductors like silicon and germanium is in between those of conductors and insulators and decreases on increasing the temperature. Semi-conductors are proving to be of great practical importance because of their marked change in conducting properties with temperature and impurity concentration.

SPECIFIC USE OF SOME CONDUCTING MATERIALS:

Tungsten: It has high melting point of 3380ºC and emits light at 2127ºC. It is thus used as a filament in bulbs.

Nichrome: It has high resistivity and melting point. It is used as an element in heating devices.

Constantan and Manganin: They have modulated resistivity. Thus they are used for making resistances and rheostats.

Tin-lead Alloy: It has low resistivity and melting point. Thus it is used as fuse wire.

1. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Solution: Given that voltage of battery V = 12 V

Circuit current I = 2.5 mA = 2.5 × 10 -3 A

∴ Value of resistance R =  V/I = 12/ 2.5 x 10 3 = 4800 Ω

2. Redraw the circuit of illustration 11, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistors. What would be the readings in the ammeter and the voltmeter?

Solution: The redrawn circuit is shown in figure. Here, ammeter A has been joined in series of the circuit and voltmeter V is joined in parallel to 12 Ω resistors.

Here total voltage of battery V = 3 × 2 = 6 V

Total resistance R = R1 + R2 + R3 = 5 + 8 + 12 = 25 Ω

∴ Ammeter reading = Current flowing in the circuit I = V/R = 6V/25 = 0.24A

∴ Voltmeter reading = Potential difference across 12 Ω resistor

V ' = IR 3 = 0.24 × 12 = 2.88 V

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How to Perform Ohm’s Law Experiment for Class 10 | Labkafe

How to perform ohm’s law experiment.

One of the most common tasks you can face in the 10th-grade physics or science laboratory is to perform the Ohm’s Law Experiment. It is common in class IX or class X in most boards like CBSE, ICSE, state boards, and even in IGCSE or IB curriculums. This experiment is also there for certain class 12 syllabus.

We have covered the   theory of Ohm’s law experiment  before, now let’s move on to the practical session of verifying Ohm’s law. But first, let’s state Ohm’s law for convenience’s sake once.

The current through a conductor between two points is directly proportional to the voltage across the two points.

Ohm’s Law Mathematical Expression

• V  = voltage across a given conductor
• I  = current through the circuit
• R  = resistance of the conductor

Base Idea of the Experiment

Here  V  would be measured across the conductor, and  I  would be measured in series to it. We will change the voltage, and record the current as a result. When we have a bunch of readings, we will plot the readings in a graph. 

Our intention is to see if the graph would come out as close to a straight line as possible. That would mean that the resistance remained the same, denoting  V  was proportional to  I . When we see that  V  and  I  are proportional, we will get the value of the resistance from the above formula.

The Experiment

To perform the Ohm’s law experiment, you will need the following equipment:

• 1 battery eliminator (0-12 volts, 2 Amps)
• 1 voltmeter 
• 1 resistance box (or an unknown resistor)
• Connecting wires

You can find all of those in Labkafe’s   Composite Lab Package  or   Physics Lab Package  .

We will need to build an electronic circuit with the battery and the resistor, through which the current is supposed to pass. To measure that current, you will need to connect the ammeter in between the battery and the resistor in  series . And to measure the voltage, you will have to connect the voltmeter in  parallel  to the resistor.

• Since the ammeter measures the current passing through the circuit, that is why it has to be a part of the circuit itself. That means, you will have to connect the positive (red) pole of the battery to the negative (black) pole of the ammeter, and connect the positive pole of the ammeter to the rest of the circuit. Since an ideal ammeter has ZERO resistance, we can assume that there will be no loss of current across the ammeter itself.

Read more:   how does an ammeter work?

• A voltmeter is supposed to measure the potential difference between two points in a circuit. Therefore, the voltmeter needs to touch those two points simultaneously with its two points, making a parallel circuit with that part of the system. Since an ideal voltmeter has INFINITE resistance, it will eat up next to no current across itself, making sure it does not influence the system.

Read more:   working principles of a voltmeter

Ideally, we should be using a battery that can give different voltages that our experiment needs. But no batteries in reality do that. So, we will put a rheostat in series with the battery, which then we can use to control the voltage as we wish. 

• A rheostat is also a resistor, just one that varies in power. So, you may naturally ask how a resistor is controlling voltage while they are supposed to control the current? The answer is quite simple ‒ just look again at the Ohm’s law formula. In “ V = IR ” we take  R  constant. But for a given, fixed value of current (keeping  I  constant), if we change the resistance value (varying  R ), the voltage  V  will change. This is a clever manipulation of Ohm’s law principle.

Even though our battery eliminator (which we are using instead of a real battery) can supply different amounts of voltage, we will keep it fixed at a constant voltage and use the rheostat to change the voltage. This will be as close to real-life as possible.

The circuit diagram for the Ohm’s law experiment is like the following:

• A is the ammeter
• V is the voltmeter
• B is the battery
• Rh is the rheostat
• R is the resistor
• B’ is the virtual, variable battery we are creating by combining B and Rh 
• S is a switch turning the circuit on and off (you may already have a switch in your battery eliminator to serve the same purpose)

(Note: the diagram in your textbook may differ somewhat from our diagram. But don’t worry, it will work out fine for our purpose.)

How to perform the experiment

• Connect the equipment as shown in the circuit diagram. Keep the switch off for now and the rheostat at the least position.
• Take a notebook and make two columns for voltage ( V ) and current ( I ). 
• Turn on the switch.
• Observe the voltmeter and ammeter carefully. In the notebook, note down the values shown in the meters in the respective columns for  V  and  I . 
• Turn off the power and move the rheostat to change its resistance a little. This will change the reading in the voltmeter.
• Repeat steps 3, 4 and 5 a few more times till you get at least half a dozen different readings.
• Plot the V vs I graph in a fresh graph paper. What does the graph look like? 

Interpreting the Results

We have performed the Ohm’s law experiment with our electronics lab package, and here is the result data as chart and graph:

As you can see, the above data shows that the current rose steadily as the voltage increased and we had a pretty much straight lined graph. That means  I  was proportional to  V . Proved!

Finding the resistance of the conductor

Since  V = IR ,

So,  R = V/I

From the above results, we can figure the average voltage as, V =  3.96v

And the average current was, I = 0.203A

Therefore, the effective value of the resistor in the experiment would be: 

R = V/I = 3.96/0.203 =  19.53Ω .

How accurate are the results of Ohm’s Law experiment?

In reality, we used a 19 ohms resistor for the experiment. The error (0.53 ohms) probably came due to: 

• Heating issues during the Ohm’s law experiment
• Internal resistances of the voltmeter, ammeter, or other components
• The equipment being used not precisely perfect to theory

Do keep in mind that due to various natural issues it would be hard for you to get an accurate value of the resistor ‒ there can probably be a ~10% error margin. The older your equipment is, the more error you can expect.

All the equipment used in this experiment came from Labkafe’s standard composite lab package.

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• Science CBSE
• Electricity
• Electric circuits - Components, effects and applications

9. Experimental verification of Ohm's law

• Connect the cell, nichrome wire, ammeter and key in series, and the voltmeter parallel to the circuit.
• First, connect only one cell in the circuit.
• Note down the current value in the ammeter and the potential difference across the nichrome wire \(XY\) in the voltmeter.
• Now, connect two cells in the circuit.
• Again, connect three cells in the circuit and note the current and the potential difference.
• Repeat the same steps using four cells in the circuit.
• Tabulate the values of current (\(I\)) and the potential difference (\(V\)) as shown in the table.

CBSE Class 10 Science Practical Skills – Ohm’s Law

BASIC BUILDING CONCEPTS

Circuit: A closed conducting loop in which electric current flows continuously is called an electric circuit or simple circuit.

You can also download Science Class 10 to help you to revise complete syllabus and score more marks in your examinations.

If net charge ‘q’ flows across any cross-sectional area in a time ‘t’, then the current flows through the conductor is

The SI unit of current is ampere(A). Flow of one coulomb of charge per second is called one ampere, i.e.

1 A = 1 coulomb per second

Small units of current are 1 mA =10 -3  A                   (mA = milliampere) 1 μA =10 -6  A                     (pA = microampere)

Direction of current: The charge carriers in metallic conductor are free electrons. The motion of these free electrons in a particular direction constitutes an electric current. Conventionally, the direction of electric current is taken as opposite to the direction of the flow of electrons. In an electric circuit, the direction of conventional current is always taken from the positive terminal of the cell/battery through the various components and then to the negative terminal.

Potential: The work done in bringing a unit positive charge from infinity to a point in an electric field is called potential at that point.

Potential difference: The potential difference between two points in an electric field or across the ends of a conductor is equal to the work done in bringing a charge from one point to another. If W is the work done in bringing a charge q from one point to another, then the potential difference between these two points is given by

V=Work Done (W)/Charge(q)

The unit of work done is joule and that of charge is coulomb. Therefore, the unit of potential difference is volt (V).

1 volt=1 joule(J)/1 coulomb(C) or 1V = 1 JC -1

Resistance: It is the characteristic property of a conducting wire which resists the flow of electric current through it. From Ohm’s law, V = IR           or              R = V/I Thus, the ratio of potential difference to the current flowing through the conductor is equal to the resistance of the conductor. The flow of electrons is retarded by the resistance of the conductor. It is expressed in volt/ampere or ohm(Ω). One ohm is equal to the resistance of a conductor through which a current of one ampere flows when one volt potential difference is applied across its ends. The resistance of a conductor depends on its

•  length,
•  cross-sectional area and
•  material of the conductor.

1 Ω = 1 VA -1

DESCRIPTION OF APPARATUS USED

AIM To study the dependence of potential difference (V) across a resistor on the current (/) passing through it and determine its resistance. Also plot a graph between V and I.

MATERIALS REQUIRED Nichrome or manganin wire, ammeter, voltmeter, battery eliminator, rheostat, one-way plug key and connecting wires.

•  Set up the circuit arrangement as shown in circuit diagram or apparatus arrangement.
•  Note the least count of the ammeter and voltmeter.
•  Find, if any, zero error of the ammeter and voltmeter and record it in table ‘A’.
•  Plug the key/switch on the battery eliminator and adjust the rheostat by sliding its variable terminal till the ammeter and the voltmeter show a reading.
• Note the readings of ammeter and voltmeter in table ‘B\ Take out the plug or switch off the battery eliminator for a moment.
•  Repeat the step 4 and 5 for the different values of current by varying the sliding terminal of rheostat.
• Tabulate all the observations in the observation table ‘B’ and find the ratio of V/I for each set of observations. Find the mean value of R.
• Plot a graph by taking / along y-axis and V along v-axis or V along .r-axis and I along y-axis.

RESULT 

• Straight line nature of the I-V graph or V-I graph shows that potential difference across the conductor is directly proportional to the current flowing through it, i.e., V x /. This proves the Ohm’s law graphically.
• The resistance of nichrome (or manganin) wire obtained from the graph is equal (or approximately equal) to the mean calculated value of R. It also verifies the Ohm’s law.
• The resistance of the given wire =………..Ω

PRECAUTIONS

•  The ends of connecting wire should be neat and clean.
•  All connections should be kept tight.
•  Positive terminal of ammeter or voltmeter should be connected to positive terminal of the battery or battery eliminator.
• The ends of resistance wire must be connected across the terminals of voltmeter.
•  Never allow the current to flow in the resistance wire for a longer time to avoid heating effect of current as R∝ T.
•  Range of voltmeter should be greater than the applied voltage.
• A low resistance ranged rheostat must be used.
•  The area of cross-section of the connecting wire should be more because it offers negligible resistance.

SOURCE OF ERROR

• Reading error may be possible while observing the pointer of ammeter and voltmeter.
• Thick connecting wires may not be available at the time of performing the experiment.
•  Area of cross-section of resistance wire may not be uniform across the length of wire.
•  There may be the use of. a high resistance rheostat.
•  Current may be allowed for a longer period of time.
•  The terminal screw of the instrument may not be tightened properly.

INTERACTIVE SESSION

Question 1: Which basic law of electricity is verified by you from this experiment? Answer: Ohm’s law.

Question 2: State Ohm’s law. Answer: If physical conditions such as temperature, pressure etc., remain the same, the current flowing through the conductor is directly proportional to the potential difference applied across it.

Question 3: In which unit current is measured? Answer: It is measured in amperes.

Question 4: What are the most essential conditions for current flow through the conductor? Answer:

• There must be a closed circuit and
• source of electrical energy.

Question 5: Which physical quantity is represented by the ratio of ‘V and T? Answer: The ratio of V and I represents the resistance of the conductor,i. e. R = V/I Question 6: What is the SI unit of resistance? Answer: The SI unit of resistance is ohm (Ω). 1Ω=1 volt/ampere.

Question 7: Why is it advised that never allow the current to flow for a long time in a resistance wire? Answer: To avoid heating effect of current. As per Joule’s law of heating effect, (H = I2Rt) the wire will be heated up which leads to increase its resistance.

Question 8: What do you mean by resistance of a conductor? Answer: The obstruction offered by the conductor to the flow of electric current through it is called resistance.

Question 9: On what factors, does the resistance of wire depend? Answer:

• The length of wire
•  Area of cross-section of the conductor
•  Nature of material

Question 10: What is least count? Answer: The least quantity that can be measured accurately by any instrument is called its least count.

Question 11: What is zero error? Answer: If the pointer of the meter (ammeter/voltmeter) does not coincide with zero of the scale, i.e., it reads a little more or less when the circuit is open, this type of error in reading of the scale is called zero error. It is always substracted from the observed reading. Question 12: There are two wires, thick and thin. Which one will have greater resistance? Answer: Thin wire as R ∝ l/A

Question 13: Give the composition of nichrome alloy. Answer: 68% Ni, 15% Cr, 15.5% Fe, 1.5% Mn. *

Question 14: What is a battery eliminator? Answer: It is used in place of cell/battery. With the help of step down transformer higher voltage of an alternating current is converted into low voltage and then converted into the direct current with the help of a rectifier.

Question 15: Why does the current not flow in the circuit when we take out the plug from key? Answer: The air present in the gap is the bad conductor of electricity. So, the circuit breaks.

NCERT LAB MANUAL QUESTIONS

Question 1: In this experiment, it is advised to take out the key from the plug when the observations are not being taken. Why? Answer: To avoid unnecessary heating of wire, it is advisable to take out the key from the plug when the observations are not being taken because current produces heating effect and the resistance increases with the increase in temperature.

Question 2: Suppose the ammeter (or voltmeter) you are using in this experiment does not have positive (+) and negative (-) terminal markings, how will you use such ammeter (or voltmeter) in the circuit? Answer: We will connect the ammeter or voltmeter arbitrary in the circuit and observe the deflection of the pointer. If the pointer strikes the stopper (i.e., before the zero reading of the scale) or deflection occurs in the opposite direction, then by interchanging the terminal connections, we can use these devices properly in the circuit.

Question 3: If the resistor of a known resistance value is replaced with a nichrome wire of say 10 cm length, how do the values of current through the nichrome wire and potential difference across the two ends of it change? How the values will change if the replaced wire is of manganin in place of nichrome? Answer: The resistivity of an alloy is generally higher than that of metals. Therefore, by replacing the known resistance value resistor with nichrome wire, the value of current through nichrome wire will decrease and potential difference across its two ends will increase. The resistivity of manganin wire is  44 x10 -8   Ωm   while that of nichrome wire is 100 x10 -8 Ωm . So, manganin wire offers less resistance as compared to the nichrome wire for the same length and same arean of cross-section. Hence, the value of current through the manganm wire will increase and the potential difference across its ends will decrease.

Question 4: Suppose in this experiment you see that the deflection on ammeter (or voltmeter) scale goes beyond the full scale. What will you infer from such an observation? What will you infer if the deflection takes place in opposite direction? Answer: If the deflection on ammeter (or voltmeter) scale goes beyond the full scale, we can infer that

•  the higher range ammeter must be used for measuring the higher value of current in the circuit.
•  the applied voltage will be very high so there will be a need of higher range voltmeter to measure the applied voltage. If deflection takes place in opposite direction, the device is not properly connected in the circuit. We must interchange the terminal connection, so that devices can be used properly in the circuit.

Question 5: Why is it advised to clean the ends of connecting wires before connecting them? Answer: To remove the insulating layer, if any, from the ends of the connecting wire.

PRACTICAL RASED QUESTIONS Multiple Choice Questions/VSA (1 Mark)

Question 2: Three students X, Y and Z, while performing the experiment to study the dependence of current on the potential difference across a resistor, connect the ammeter (A), the battery (B), the key (K) and the resistor (R) in series, in the following three different orders. X——>B, K, R, A, B Y ——> B, A, K, R, B Z ——> B, R, K, A, B Who has connected them in the correct order? (a) X (b) Y (c) Z (d) All of them

Question 4: A voltmeter has a least count of 0.05 volt. While doing Ohm’s law experiment, a student observed that the pointer of the voltmeter coincides with 15th division, the observed reading is (a) 0.75 V (6) 0.075 V (c) 7.5 V (d). 75 V

Question 5: In an experiment to study dependence of current on the potential difference across a given resistor, four students P, Q, R and S kept the plug key in the circuit closed for time t1 and then open

Question 6: The following ‘precautions’ were listed by a student in the experiment on study of ‘Dependence of current on potential difference’: [Delhi 2009] (A) Use copper wires as thin as possible for making connections. (B) All the connections should be kept tight. (C) The positive and negative terminals of the voltmeter and the ammeter should be correctly connected. (D) The ‘zero error’ in the ammeter and the voltmeter should be noted and taken into consideration while recording the measurements. (E) The ‘key’ in the circuit, once plugged in, should not be taken out till all the observations have been completed. The ‘precautions’ that need to be corrected and revised are (a) (A), (C) and (E) (b) (C) and (E) (c) (B) and (E) (d) (A) and (E)

Question 8: An ammeter has 20 divisions between mark 0 and mark 2 on its scale. The least count of the ammeter is (a) 0.02 A (6) 0.01 A (c) 0.2 A (d) 0.1 A

Question 9: In a voltmeter, there are 20 divisions between the 0 mark and 0.5 V mark. The least count of the voltmeter is (a) 0.020 V (6) 0.025 V (c) 0.050 V (d) 0.250 V

Question 12: Three students A, B and C carried out measurements of current and potential difference with the help of the ammeter and voltmeter, they did the following A: viewed the divisions from the left of the pointer B: viewed the divisions from the right of the pointer C: viewed the divisions directly above the pointer The correct procedure followed by the student (a) A (b) B (c) C (d) none of them

Question 13: The plane mirror is used in the meters is to (a) avoid the error of parallax (b) help to take more accurate observations (c) reflected light makes the scale illuminated (d) make the meter look good

(a) K (b) L (c) M (d) N

Short Answer Questions

Question 2: Name and state the law that connects the electric current flowing through a metallic resistor and potential difference across its two ends. State the condition under which it is valid.

Question 4: Mention the condition under which charges can move in a conductor. Name the device which is used to maintain this condition in an electric circuit for verifying Ohm’s law.

•  What is the relation between V and I?
•  Find the resistance of the conductor using the graph.

Multiple Choice Questions/VSA 1. (b) 2. (d) 3. (b) 4. (a) 5. (a) 6. (d) 7. (c) 8. (d) 9. (b) 10. (d) 11. (b) 12. (c) 13. (a) 14. (b) 15. (c)

2. Ohm’s Law: It states that the potential difference (V) across the ends of a given metallic conductor in an electric circuit is directly proportional to the current (I) flowing through it. Ohm’s law is valid only when the temperature of the conductor remains constant.

3. Using Ohm’s law, V= IR 3=I 3 I = 1 A Therefore, ammeter reading is 1 A. Voltmeter reading = Potential difference across 3 Q resistor V= IR = 1 3 = 3 V.

4. Electric potential should be maintained across the ends of a conductor to move the charge through it. Cell or battery is the device used to maintain the potential difference across the conductor.

Science Practical Skills Science Labs Math Labs Math Labs with Activity

• 10 Science Lab Manual
• Ohms Law Solution

Class 10 th Science Lab Manual CBSE Solution

• AIMTo study the dependence of potential difference (V) across a resistor on the current…

Define a circuit diagram?

Define electric current. Give its SI unit?

Define 1 A of current?

• What is mA and μA?

What is the direction of current in a circuit?

• What are the functions of the following components of electric circuit? i. Ammeter ii.…
• What is the difference between electric potential electric potential difference?…
• What is resistance and factors affecting the resistance of a body?…
• What is the difference between voltmeter ammeter?

How are ammeter and voltmeter connected in a circuit?

What is the ideal resistance of ammeter and voltmeter?

What is Rheostat?

State Ohms law?

• Why it is never advised to keep the current flow for a long time in a resistance wire?…
• What is an essential condition for current flow through the conductor?…
• Why does current not flow in the circuit when we take out the plug from the key?…

What is battery eliminator?

• How do you mean by slope of a graph? What is the formula for slope?…
• Which has more slope, A or B?

What is the resistance of a ideal ammeter?

Lab Experiment 4

AIM To study the dependence of potential difference (V) across a resistor on the current (I) passing through it and determine Its resistance and also plotting a graph between V and I.

MATERIALS REQUIRED

Resistance wire, ammeter, voltmeter, battery eliminator, rheostat, One-way plug key, and connecting wires.

Ohm’s Law states that “If the physical conditions such as temperature, pressure, etc., remain the same during the experiment, then the current (I) flowing is directly proportional to the potential difference (V) across the ends of the circuit.”

Mathematically,

1. The resistance(R) is the characteristic property of the conductor which resist the flow of electric current through it.

2. The potential difference (V) is the is the potential difference across the ends of a conductor.

3. The electric current (I) is the amount of charge flowing through a particular area in a unit time.

If we plot a graph between the current (I) and the applied potential difference(V) between its ends, for an ideal resistance it will be a straight line as shown:

Circuit Diagram

In the above diagram

Apparatus Arrangement :

The actual diagram of the Ohm’s Law apparatus is shown here:

1. By using the circuit diagram or apparatus arrangement, we set up the circuit for finding the dependence of voltage on the current flowing in the circuit.

2. Clean the end of connecting wires using with sandpaper to remove the insulation.

3. Determine the least count of the ammeter and voltmeter and note them under observation section.

4. Check for any zero error in the ammeter and voltmeter and if any record it in table ‘A.’

5. Switch on the battery eliminator, plug the key and adjust the resistance offered by the rheostat by sliding its variable terminal till the ammeter and the voltmeter show a reading.

6. Write the readings of ammeter and voltmeter in the observation table. Take out the key plug out of the circuit to make the circuit open.

7. Repeat the process done in step 4 and 5 for the different values of current by varying the sliding terminal of the rheostat and note down the reading for a respective value of voltage for current in the observation table.

8. Note all the observations in the observation table ‘B’ and then find the ratio of - for each set of observations. Find the mean value of R.

9. Plot a graph by taking I along the y-axis and V along the x-axis.

Observations:

OBSERVATION TABLE

1. Least count of ammeter:

The image of the ammeter is attached here:

The range of the ammeter =500 - 0 mA = 500 mA = 0.5 A

The number of divisions in between two consecutive values= 10

2. Zero error of ammeter:

The needle of the ammeter points towards zero of the main scale of the ammeter.

3. Least count of voltmeter:

The range of voltmeter = 2 - 0 V= 2 V.

The number of division in small scale between two division on the main scale

4. Zero error of voltmeter = 0 V.

Table (B) for the reading of ammeter and voltmeter

CALCULATION

1. The ratio of V and I for each corrected set of observation is given in the table:

The mean value of Resistance is calculated as:

2. The graph between the potential difference ‘V’ along the x-axis and the current ‘F along the y-axis for I-Vgraph as shown or V’ on the y-axis and ‘I’ an x-axis is drawn:

3. Find the slope of the line.

(i) For I-V graph.

:. Resistance of nichrome (Or manganin) wire =1.97 Ω.

(ii) For the V-I graph,

wire = 2.06 Ω.

1. The linear nature of the I-V graph or V-I graph shows that potential difference across the end of the conductor is directly proportional to the current flowing through it, i.e., V ∝ I. This proves the Ohm’s law graphically.

2. The resistance of the wire obtained from the graph is equal to the mean calculated value of R. It also verifies the Ohm’s law.

3. The resistance of the given wire = 2.05 Ω.

PRECAUTIONS

1. The ends of connecting wire should be rubbed using sandpaper.

2. All connections should be kept clean and tight.

3. The positive terminal of ammeter or voltmeter should be connected to the positive terminal of the battery or battery eliminator.

4. The ends of the resistance wire must be connected across the terminals of the voltmeter.

5. Remove the key out of the circuit when not in use to avoid the heating effect of the circuit.

Viva Questions

The circuit diagram is a schematic representation which shows the arrangement of different devices or components by using their electrical symbol is called a circuit diagram.

The basic circuit diagram is given as:

Electric current: It is the amount of charge following through a particular area in a unit time is called an electric current.

The SI unit of current is 1A.

The 1A of current is the flow of 1 coulomb of charge per second.

What is mA and μA?

Ampere is the standard unit of current. The small units of current are:

1. Milli-Ampere: 1 mA = 10 -3 A

2. Micro-Ampere: 1Μa = 10 -6 A

The conventional direction of current is taken from positive terminal of the battery through the various circuit components & then to negative terminal.

What are the functions of the following components of electric circuit? i. Ammeter ii. Voltmeter iii. Resistor iv. Switch v. Battery

The functions of the components are given as:

i. Ammeter: It is a measuring instrument used to measure the current in a circuit.

ii. Voltmeter: A voltmeter is an instrument used for measuring electrical potential difference between two points in an electric circuit

iii. Resistor: It is an electrical component that limits or regulates the flow of electrical current in an electronic circuit.

iv. Switch: A device for making and breaking the connection in an electric circuit.

v. Battery: A battery is an electrochemical cell that can be charged electrically to provide a static potential for power.

What is the difference between electric potential & electric potential difference?

i. The work done in bringing a unit positive charge from infinity to a point in an electric field is called as Electric potential of a point.

ii. The potential difference between two points in an electric field or across the ends of a conductor is equal to the work done in bringing a charge from one point to another

What is resistance and factors affecting the resistance of a body?

Resistance: It is the characteristic property of a conducting wire which resists the flow of electric current through it. Its SI unit is ohms.

The factor affecting the resistance of a body are:

1. Length of the wire.

2. Cross-sectional area.

3. The material of the conductor.

4. The temperature of the conductor.

What is the difference between voltmeter & ammeter?

Voltmeter is a very high resistance device which is used to measure the potential difference between two points. It is connected in parallel combination with the resistance across which the voltage drop has to be found.

It is a very low resistance device which is used to measure the strength of the current in a circuit, it is always connected in series in a circuit.

The ideal resistance of voltmeter is infinite ohms.

The ideal resistance of ammeter is 0 ohms.

If the physical conditions such as temperature, pressure remain the same, the current flowing through a conductor is directly proportional to the potential difference applied across it. Mathematically, V = IR

Where V is the potential difference I is the current

R is the resistance

Why it is never advised to keep the current flow for a long time in a resistance wire?

The flowing current in the circuit led to the heating effect of current. As per joule law H = I 2 RT, the wire will be heated up which increases the resistance of wire.

What is an essential condition for current flow through the conductor?

The necessary condition are:

1. The circuit must be closed.

2. There must be a source of electrical energy.

Why does current not flow in the circuit when we take out the plug from the key?

Where there is no connectivity in the medium, the current does not flow. The air present in the gap caused due to removal of plug from the key is the bad conductor of electricity. Hence, the current does no flow.

It is used in place of cell/ battery. With the help of step down transformer higher voltage of an alternating current is converted into a low voltage and then converted into direct current with the help of a rectifier.

How do you mean by slope of a graph? What is the formula for slope?

The slope is a measure of the steepness of any line connecting two points.

For a line connecting any two points say P(x 1 , y 1 ) and Q (x 2 , y 2 ), the slope is given as :

The resistance of an ideal ammeter is zero.

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Ohm’s Law Physics Lab Manual Class 10

Free pdf download.

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There is no mode of learning better than learning from practical knowledge and that is why CBSE has prescribed students the Lab Manual of Class 10 Physics Ohm’s Law. Students looking for an individual file of Ohm’s Law Physics Lab Manual Class 10 can use the link we have provided here on this page.

Interestingly, the PDF file not only contains the process to perform an activity on the Ohm’s Law but helps students prepare for the internal assessment and Viva examination.

What Includes Ohm’s Law Lab Manual Class 10 PDF?

Inside the Ohm’s Law Lab Manual Class 10 PDF, there are various helpful and important things for students - Discussed below in detail:

• Quick Revision Notes: To quickly recap the concepts of the Ohm’s Law the PDF file of CBSE Class 10 Lab Manual Physics contains short notes to refresh the learning of students before performing the activity.
• Aim: The purpose of the Ohm’s Law activity is mentioned in the Aim section. It helps students understand what should be the final result of this laboratory activity.
• Materials Required: All the apparatus required to perform the practical activity on the Ohm’s Law is mentioned in this section.
• Theory: The theory section explains the Ohm’s Law in a little short brief to give students an initial idea of the topic to perform the activity given in the lab manual.
• Graphics/Images/Illustrations: To help students better understand the process or some of the apparatus, the Ohm’s Law Physics Lab Manual Class 10 contains images, graphs and illustrations.
• Procedure: This section includes, a stepwise guide to help students to perform the activity on the Ohm’s Law.
• Observations: The observed data extracted from performing the activity on the Ohm’s Law is mentioned in this section.
• Precautions: One of the crucial things while conducting the activity given in Class 10 Ohm’s Law lab manual is to take precautions to avoid making mistakes or getting hurt. The precautions section in the lab manual explains students to what to not do during the activity of Ohm’s Law.
• Source of Error: The PDF that we provide contains, the source of error that help students understand the cause of not getting the desired outcome.

Where to Download Lab Manual of CBSE Class 10 Ohm’s Law?

Always choose a trustworthy platform to download study materials such as Selfstudys.com. If you want to download the PDF file of Ohm’s Law from Selfstudys then follow the steps given below:

• Navigate to the website (Selfstudys.com) on any internet browser
• Once, the website loads, tap on the navigation button

• Tapping on the navigation button will give you further options to choose from - Again, tap on CBSE

• In this step, you are required to tap or click on Lab Manual

• A new page will open, choose Class 10 and then click on Physics to access the PDF file of Ohm’s Law Lab Manual of CBSE Class 10.

What is the Significance of the Class 10 Ohm’s Law Lab Manual?

From helping students perform the practical activity to preparing for the annual Viva examination, the Class 10 Ohm’s Law Lab Manual plays a significant role in a student’s academic session. Here are some of the key significant roles:

• Helps Prepare Practical Notebook: For internal assessment, students must prepare the practical notebook of Class 10 Physics and while preparing the practical notebook the Class 10 Ohm’s Law Lab Manual PDF can help students. It can help because it works like a guidebook.
• Assistance in Practising Viva Questions: The PDF file of the Class 10 Ohm’s Law Lab Manual we provide contains Viva questions along with some extra MCQ types of questions. Thus, here, the Ohm’s Law Lab Manual helps students practice Viva questions to be ready for the upcoming annual examination.
• Helps Develop a Better Command of Class 10 Ohm’s Law: By helping students practice activities mentioned in the Class 10 Ohm’s Law Lab Manual, it enables them to develop a better and stronger command of the topics and concepts covered.

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Learn all about Ohm's law

Ohm’s law experiment

Today you’ll learn a step by step guide to perform the  Ohm’s law Experiment . You’ll learn the use of voltmeter and ammeter in parallel and series, resistors, dc power supply, wires and all other equipment which is used in doing the practical.

This article demonstrates the Ohm’s Practical experiment. You can find the  lab report, reading, observations, and theory here .

Steps to Perform Ohm’s Law Experiment

• Connect the resistor on the breadboard.
• Connect the source to the breadboard.
• Connect the ammeter in series.
• Connect the voltmeter in parallel.
• Increase the voltage step-by-step from 0 to10 V and note the voltage/current.

Let’s start with the  circuit diagram :

The graphic below illustrates a visual representation of steps:

Resistor Color Coding

Carbon resistors are the most popular type of resistors which are used in lab experiments. Circular color bands are used on their bodies are to indicate the amount of resistance they possess. We can use a color coding chart to find the value of carbon resistors. Let’s consider your resistor box has three resistors.

Resistor 1 is a 4 Band Resistor with Brown, Black, Red, Gold bands on it.

Resistor 2 has Red, Red, Red, Gold bands on it.

Resistor 3 has Red, Red, Orange, Gold bands on it.

The first band of all resistors is the first digit of resistance, the second band is the second digit of resistance. The third band indicates the multiplier values whereas the fourth band indicates the tolerance value of resistors.

Let’s understand how a color coding chart helps us to find the value of resistance.

Let’s decode our first resistor:

Here the first band is brown so the first digit is 1, the second band is black so second digital 0, the third red band provides 100 as multiplier which is 10 * 100 = 1000 Ω. The last golden band represents a tolerance of 1 kΩ ± 5%.

The second and third resistors are decoded to be 2.2 kΩ ± 5% and 22 kΩ ± 5%.

Variable DC Power Supply

A variable dc power supply or VDC has a circular knob which can be rotated to obtain variable voltages. An LCD displays the number of selected output voltages. The alligator wires are used to take the output from the supply. Always connected red wire/probe to red (live) terminal of the battery and black wire to the black terminal of the battery. The figure below displays a variable dc power supply. While performing the experiment you rotate the knob to obtain step by step variable voltages.

A Voltmeter is connected to the resistor for measuring the amount of potential difference across its ends. A voltmeter always connects in parallel because the potential difference remains same in parallel. Practically voltmeter, ammeter, and ohmmeter are designed together in the form of the multimeter. So you need to set your multimeter to the voltage scale.

An ammeter is a current measuring device. It connects in series to the circuit under measurement.

• Ohm’s law circuit diagram [How to Read Symbols] →

• CBSE Class 12
• CBSE Class 12 Physics Practical
• Determine Resistance Plotting Graph Potential Difference Versus Current

To Determine Resistance per CM of a Given Wire by Plotting a Graph for Potential Difference versus Current

We know that resistance is a measure of the opposition to the flow of electricity in a circuit. The potential difference helps to understand the amount of energy transferred between two points in a circuit. In this session, let us learn to determine the resistance per cm of a given wire by plotting a graph for potential difference versus current.

To determine the resistance per cm of a given wire by plotting a graph for potential difference versus current.

Apparatus/Material Required

• A wire of unknown resistance
• Milliammeter
• Connecting wires
• Piece of sandpaper

Circuit Diagram

According to Ohm’s law, the electric current flowing through a conductor is directly proportional to the potential difference across its ends, provided the physical state (pressure, temperature, and dimensions) of the conductor remains unchanged.

If I is the current flowing through the conductor and V is the potential difference across its end, then

Where R is the constant of proportionality and is termed as the electrical resistance of the conductor. Resistance R depends on the dimensions and material of the conductor. The relationship between the resistance of a material and its length and area of the cross-section is given by the formula

Where ρ is the specific resistance or resistivity and is a characteristic of the material of the wire.

• Clean the ends of the connecting wire with the help of sandpaper to remove any insulating coating on them.
• Connect the resistance, rheostat, battery, key, voltmeter, and ammeter as shown in the figure.
• Make sure that the pointers in the voltmeter and milliammeter coincide with the zero mark on the measuring scale. If not, adjust the pointer to coincide with the zero mark by adjusting the screw provided at the base using a screwdriver.
• Note the range and the least count of the given voltmeter and milliammeter.
• Insert the key K and slide the rheostat to the end where the current flow is minimum.
• Note the voltmeter and the milliammeter reading.
• Remove the key K and allow the wire to cool. Again insert the key and slightly increase the voltage by moving the rheostat. Note down the milliammeter and voltmeter reading.
• Repeat step 7 for four different adjustments of the rheostat. Document the readings in a tabular column.

Observations

Range of ammeters = _____ mA to _____ mA

The least count of ammeter = _____ mA

Range of voltmeter = _____ V to ____ V

The least count of voltmeter = _____ V

The least count of meter-scale = _____ m

Length of the given wire, l = _____ m

Calculations

• Plot a graph between the potential difference across the wire V and the current I flowing through the wire as shown below.

2. Determine the slope of the graph. The resistance of the given wire is then equal to the reciprocal of the slope.

From the graph, R = BC / AB = _____ Ω

3. Resistance per unit length of the wire = R/t = _____ Ωm –1

Here, R is the resistance per unit length and Δ R is the estimated error. Δ V  and Δ I are the least count of voltmeter and ammeter respectively.

The potential difference across the wire varies linearly with the current.

The resistance per unit length of the wire is ( R ± Δ R ) = _____ ± _____ Ωm –1 ).

1. State Ohm’s Law.

Ohm’s law states that the potential difference across an ideal conductor is proportional to the current through it. The constant of proportional is known as the resistance R . Ohm’s law is given by V = IR .

2. Which are the factors on which the resistance of a conductor depends on?

The resistance of a conductor depends on the following factors:

• Resistivity
• Temperature
• Cross-sectional area

3. What is a rheostat?

A rheostat is a variable resistance that is used to control the current.

4. What is the shape of a V vs I graph for a linear resistor?

The shape of the V vs I graph for a linear resistor is a straight line.

5. What is the reciprocal of resistivity called?

The reciprocal of resistivity is called conductivity.

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