The Figure 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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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,...
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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 withR= 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. read less
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