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 :

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

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 :

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

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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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3/4 times the value after closing S₁

3/2 times the value after closing S₁

3/2 times the initial value

3/4 times the initial value

answer is B.

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Detailed Solution

Initially : V_V + V_A = 6
V_V and V_A being the potential across voltmeter and ammeter respectively. After closing S₁,
$\frac{V_{V}}{2} + 2 V_{A} = 6 \Rightarrow V_{V} = 4, V_{A} = 2$ after closing $S_{2} : V_{V} = 0, V_{A} = 6$
So that value after closing S₂ is 3/2 times the value after closing S₁