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Connect a 1.5 V dry cell to a resistor (low resistance), an ammeter and a key in series as shown in Fig. Record the electric current shown by the ammeter. Replace the single cell by 2, 3, … cells.

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Question

Connect a 1.5 V dry cell to a resistor (low resistance), an ammeter and a key in series as shown in Fig. Record the electric current shown by the ammeter. Replace the single cell by 2, 3, … cells. For the number of cells, record the electric current as is done for you in the table given below: 

S.No. P.d. (V)
(in volts)
Electric current
(in amperes)
V / I
1. 1.5 3 ...
2. 3.0 6 ...
3. 4.5 9 ...
4. ... ... ...

The figure shows that increasing the number of cells increases the potential difference across a conductor, causing more electric current to flow. Similarly, water flow increases with greater height or pressure.

Activity
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Solution

Aim:

To verify Ohm’s law, i.e., to show that the electric current flowing through a conductor is directly proportional to the potential difference across its ends, at constant temperature.

Apparatus Required:

A dry cell (1.5 V), a low-resistance resistor, an ammeter, a key (switch), connecting wires and a voltmeter.

Procedure:

  1. Connect a single 1.5 V dry cell, the low-resistance resistor, the ammeter and the key in series as shown in Fig.
  2. Close the key and record the ammeter reading (electric current) and the potential difference across the resistor.
  3. Replace the single cell by 2 cells, then 3 cells, then 4 cells, and so on.
  4. Each time note the potential difference (p.d.) and the corresponding electric current.
  5. Compute \[ V/I \] for each observation and record it in the table.

Observation Table:

S.No. P.d. (V) (in volts) Electric current (I) (in amperes) (V / I) \((\Omega)\)
1. 1.5 3 \[0.5\ \Omega\]
2. 3.0 6 \[0.5\ \Omega\]
3. 4.5 9 \[0.5\ \Omega\]
4. 6.0 12 \[0.5\ \Omega\]

Calculation of (V/I) in Each Case:

\[ \frac{V}{I} = \frac{1.5}{3} = 0.5\ \Omega \] 

\[ \frac{V}{I} = \frac{3.0}{6} = 0.5\ \Omega \] 

\[ \frac{V}{I} = \frac{4.5}{9} = 0.5\ \Omega \] 

\[ \frac{V}{I} = \frac{6.0}{12} = 0.5\ \Omega \]

Observations and Conclusion:

  1. The ratio \[V/I\] remains constant (equal to \(0.5\ \Omega)\) in every observation, even though (V) and (I) both change.
  2. When the p.d. is doubled (from 1.5 V to 3.0 V), the current also doubles (from 3 A to 6 A). When the p.d. becomes three times, the current also becomes three times.
  3. Hence, \[\ I \propto V \] or \[\ V \propto I \], which verifies Ohm’s law.
  4. A graph of (I) against (V) is a straight line passing through the origin, confirming a direct relationship between current and potential difference.

Precautions:

  1. The key should be closed only while taking the readings to avoid unnecessary heating of the resistor.
  2. The ammeter must be connected in series and the voltmeter in parallel across the resistor.
  3. The temperature of the conductor should remain constant throughout the experiment.
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Chapter 4: Electricity and Magnetism - Activity 4.1 [Page 223]

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Frank Physics Part 2 [English] Class 10 ICSE
Chapter 4 Electricity and Magnetism
Activity 4.1 | Q 1. | Page 223
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