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Question

A parallel plate capacitor of 1 μF capacity is discharging through a resistor. If its energy reduces to half in one second. The value of resistance will be

A
2ln(2) MΩ
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B
4ln(2) MΩ
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C
10ln(2) MΩ
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D
16ln(2) MΩ
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Solution

The correct option is A 2ln(2) MΩ
As Q=Qoet/τ
When energy is 50% then Q=Qo2
Qo2=Qoet/τ
et/τ=2
tτ=ln(2)
τ=tln(2)
RC=2ln2
R=2Cln2=2106×ln2
=2×106ln2 Ω

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