Question 705043

SCQEASY

A battery with an open circuit voltage of $10 V$ is connected to a load resistor of $485\Omega$ and the voltage measured across the battery terminals using an ideal voltmeter is $9.7 V$ . The internal resistance of the battery is closest to

(A)

30\Omega

(B)

15\Omega

(C)

20\Omega

(D)

40\Omega

Detailed Solution

The load current is $I = \frac{9.7 V}{485 \Omega} = 20 mA$ . The voltage drop across the internal resistance $r$ is $10 V - 9.7 V = 0.3 V$ . Thus, $r = \frac{0.3}{20 \times 10^{-3}} = 15 \Omega$ .

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