Question
Answer: Option C
:
C
In the steady state no current flows in the branch containing the capacitor .Thus the current say l flows in the branches containing R and 2R. Applying Kirchoff's second rule to the loop abcdefa
2V-I(2R)-IR-V=0
l=V3R
Potential drop across capacitor =2V-V-l(2R)=V−v3R×2R
=V−2v3=v3
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:
C
In the steady state no current flows in the branch containing the capacitor .Thus the current say l flows in the branches containing R and 2R. Applying Kirchoff's second rule to the loop abcdefa
2V-I(2R)-IR-V=0
l=V3R
Potential drop across capacitor =2V-V-l(2R)=V−v3R×2R
=V−2v3=v3
Was this answer helpful ?
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