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Question

Find the charge and potential drop across the capacitor C4 in the given circuit ?


A
407 μC, 207 V
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B
807 μC, 407 V
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C
307 μC, 157 V
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D
207 μC, 407 V
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Solution

The correct option is A 407 μC, 207 V
We can solve this by potential method, by assigning potential as shown in the figure.


Using KVL for isolated system I,

(V15)4+(V10)2+(V1V2)4=0

4V120+2V1+4V14V2=0

10V14V2=20 ....(1)

Using KVL for isolated system II,

(V25)2+(V20)4+(V2V1)4=0

2V210+4V2+4V24V1=0

10V24V1=10 ...(2)

Solving eqn (1) and (2)

V1=207 V and

V2=157 V

Charge and potential drop across capacitor C4 is,

Q4=C4(V10)=2×(2070)=407 μC

V4=Q4C4=4072=207 V

Hence, option (a) is the correct answer.

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