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  • In the circuit shown, the battery is ideal, with emf E = 15 V and it sends a current I in the circuit. All resistors are identical and each resistor has resistance R = 3Ω. The potential

In the circuit shown, the battery is ideal, with emf E = 15 V and it sends a current I in the circuit. All resistors are identical and each resistor has resistance R = 3Ω. The potential difference (in V)  across the capacitor in steady state is Vc = 

Option: 1

12


Option: 2

9


Option: 3

0


Option: 4

15


Answers (1)

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In steady state, no current passes through the branch that contains a fully charged capacitor, because a fully charged capacitor is a dc blocking element. Hence the circuit becomes 

\begin{array}{l}{\text { For Loop AGFEDCA}} \\ \\ {-2 \mathrm{I}_{1} \mathrm{R}-\mathrm{IR}+\mathrm{E}=0} \\ {\Rightarrow \quad 6 \mathrm{I}_{1}+3 \mathrm{I}=15} \dots(1) \\ {\text { For Loop BECDB} }\\ { \left(\mathrm{I}_{1}-\mathrm{I}\right) \mathrm{R}-\mathrm{IR}+\mathrm{E}=0} \\ {-2 \mathrm{IR}+\mathrm{I}_{1} \mathrm{R}+\mathrm{E}=0} \\ {\Rightarrow \quad 3 \mathrm{I}_{1}-6 \mathrm{I}=-15} \\ {\Rightarrow \quad \frac{3 \mathrm{I}_{1}}{2}-3 \mathrm{I}=-\frac{15}{2} \dots(2)}\end{array}

\begin{array}{l}{\text { Add }(1) \text { and }(2), \text { we get }} \\ {\qquad\left(\frac{3}{2}+6\right) I_{1}=15-\frac{15}{2}} \\ {\Rightarrow \quad\left(\frac{3+12}{2}\right) I_{1}=\frac{30-15}{2}} \\ {\Rightarrow \quad 15 I_{1}=15} \\ {\Rightarrow I_{1}=1 A} \\ {\Rightarrow \quad 6+3 I=15} \\ {\Rightarrow 31=9} \\ {\Rightarrow I=3 \mathrm{A}}\end{array}

\begin{array}{l}{\text { For Loop GHEFG}} \\ {\mathrm{V}_{\mathrm{C}}+\mathrm{I}(3)+\mathrm{I}_{1}(3)=0} \\ {\Rightarrow \quad \mathrm{V}_{\mathrm{C}}=9+3=12 \mathrm{V}}\end{array}

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Pankaj

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