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Question

A charged capacitor is allowed to discharge through a resistor by closing the key at the instant t=0. At J the instant t=(ln4)μs, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to

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A
0.5Ω
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B
1Ω
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C
2Ω
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D
4Ω
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Solution

The correct option is C 2Ω
During discharging, current , i(t)=i0et/RC

If r be the resistance of ammeter, net resistance , R=2+r

When t=ln4,i=i0/2i0e(ln4/RC)=i0/2

or e(ln4/RC)=2

take ln on both sides, ln4/RC=ln2

or RC=ln4ln2=2ln2ln2=2

or R=2C=2/0.5=4

or 2+r=4r=2Ω

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