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When an ammeter of negligible internal resistance is inserted in series with the given circuit, it reads 1 A. When the voltmeter of very large resistance is connected across X, it reads 1 V. When the point A and B are shorted by a conducting wire, the voltmeters measures l0 V across
the battery. The internal resistance of the battery is equal to

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zero

$0.2 Ω$

$0.1 Ω$

$0.5 Ω$

answer is C.

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Detailed Solution

$\frac{12}{X + Y + r} = 1 A; Also \Rightarrow V_{X} = 1 = 1 \times X \Rightarrow X = 1 Ω$

When Y is shorted, $I = \frac{12}{1 + r}$

$\begin{array}{l} 10 = 12 - Ir \Rightarrow 10 = 12 - \frac{12 r}{{(1 + r)}^{}} \\ \Rightarrow 10 + 10 r = 12 + 12 r - 12 r \\ \Rightarrow 10 r = 2 \Rightarrow r = 0 .2 Ω \end{array}$

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