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प्रश्न
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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उत्तर
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,
r = `(("l"_1 - "l"_2)/"l"_2) "R"`
= `(76.3 - 64.8)/64.8 xx 9.5`
= 1.68 Ω
Therefore, the internal resistance of the cell is 1.68 Ω.
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