A parallel plate capacitor of 1μF 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 A2ln(2)MΩ As Q=Qoe−t/τ
When energy is 50% then Q=Qo√2 Qo√2=Qoe−t/τ ⇒et/τ=√2 tτ=ln(√2) τ=tln(√2) RC=2ln2 R=2Cln2=210−6×ln2 =2×106ln2Ω