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Question

A capacitor is connected to a 10 V battery through a resistance of 20 Ω. It is found that the voltage across the capacitor rises to 8 V in 1 μs. The capacitance of the capacitor is approximately equal to [Take ln5=1.6]

A
4×108 F
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B
3×108 F
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C
2×108 F
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D
6×108 F
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Solution

The correct option is B 3×108 F


The charge on the capacitor during charging is given by,

Q=Q01etRC

Hence, voltage across the capacitor at any time t will be,

V=QC=Q0C1etRC

Here, at t=1 μs, V=8 V whereas the steady state voltage will be (Q0C=10 V) when the capacitor is fully charged.

8=101etRC

etRC=15

tRC=ln5=1.6

As, t=1 μs=106 s and R=20 Ω

C=1061.6×203×108 F

Hence, option (b) is the correct answer.

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