Q.

An ammeter and a voltmeter are initially connected in series to a battery of zero internal resistance. When switch is closed the reading S1 of the voltmeter becomes half of the initial, whereas the reading of the ammeter becomes double. If now switch S2 is also closed, then reading of ammeter becomes :

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a

3/4 times the value after closing S1

b

3/2 times the value after closing S1

c

3/2 times the initial value

d

3/4 times the initial value

answer is B.

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Detailed Solution

Initially : VV + VA = 6
VV and VA being the potential across voltmeter and ammeter respectively. After closing S1,
VV2+2VA=6VV=4,VA=2 after closing S2:VV=0,VA=6
So that value after closing S2 is 3/2 times the value after closing S1

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