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When an ammeter of negligible internal resistance is inserted in series with 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 voltmeter measures 10 V. The internal resistance of the battery (r) 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

I = 1 A

r is the internal resistance of the battery

$12 = (X + Y + r) (1) .............. (i)$

Also $1 = (X) (1) o r X = 1 Ω$

Voltage across X (when A and B are shorted)

$10 = (\frac{12}{X + r}) X o r 10 = \frac{12}{1 + r} o r 1 + r = \frac{6}{5}$

$o r r = \frac{1}{5} Ω$

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