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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 VWe can solve this by potential method, by assigning potential as shown in the figure. Using KVL for isolated system −I, (V1−5)4+(V1−0)2+(V1−V2)4=0 ⇒4V1−20+2V1+4V1−4V2=0 ⇒10V1−4V2=20 ....(1) Using KVL for isolated system −II, (V2−5)2+(V2−0)4+(V2−V1)4=0 ⇒2V2−10+4V2+4V2−4V1=0 ⇒10V2−4V1=10 ...(2) Solving eqn (1) and (2) V1=207 V and V2=157 V Charge and potential drop across capacitor C4 is, Q4=C4(V1−0)=2×(207−0)=407 μC V4=Q4C4=4072=207 V Hence, option (a) is the correct answer.

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