Figure 3-35 shows a 2-0 V potentiometer used for the determination of internal resistance of a 1-5 V cell- The balance point of the cell in open circuit is 76-3 cm- When a resistor of 9-5 - is used in the external circuit of the cell- the balance point shifts to 64-8 cm length of the potentiometer wire- Determine the internal resistance of the cell-

Internal resistance of the cell = r Balance point of the cell in open circuit, l1 = 76.3 cm An external resistance (R) is connected to the circuit with R = 9.5 ...

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Standard XII

Physics

Forward Bias

Figure 3.35 s...

Question

Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

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Solution

Internal resistance of the cell = r

Balance point of the cell in open circuit, l₁ = 76.3 cm

An external resistance (R) is connected to the circuit with R = 9.5 Ω

New balance point of the circuit, l₂= 64.8 cm

Current flowing through the circuit = I

The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1.68Ω.

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