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Question

In the given circuit, find ratio of charges on 4 μF and 2 μF capacitors in steady state.


A
43
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B
65
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C
49
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D
79
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Solution

The correct option is C 49

In the given circuit let current I in the single loop be as shown in figure, as we know that in steady state , no current flows through the branch containing capacitor, so we do not consider any current in the branches with capacitor as shown.



Current in the single loop AHCDEFGA is,

I=2010(1+1+1+2)=2 A

Using KVL to the open loop (AHCB) gives,

VA(2×1)(2×1)=VB

VAVB=4 V

Using KVL to the open loop (HCDE) gives,

VH(2×1)+20=VE

VHVE=18 V

Charges on the given 4 μF and 2 μFcapacitors as follows,

q4 μ F=CVAB

q4 μ F=4×4=16 μC

q2 μ F=CVEH

q4μF=2×18=36 μC

q4 μ Fq2 μ F=1636=49

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.

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