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  • .In the given circuit, with steady current, the potential drop across the capacitor must be <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/20/f52_q127.PNG" style="heig

.In the given circuit, with steady current, the potential drop across the capacitor must be

Option: 1

\mathrm{V}


Option: 2

\mathrm{\frac{V}{2}}


Option: 3

\mathrm{\frac{V}{3}}


Option: 4

\mathrm{\frac{2V}{3}}


Answers (1)

best_answer

Method I: In this, there will be no current flowing in branch B E in steady condition. Let I be the current flowing in the loop ABCDEFA. Applying Kirchhoff's law in the loop moving in anticlockwise direction starting from \mathrm{C+2 V-I(2 R)-I(R)-V=0 \Rightarrow V= 3 I R \Rightarrow I=V / 3 R} Applying Kirchhoff's law in the circuit A B E F / A we get on moving in anticlockwise direction starting from B

\mathrm{+V+V_{\text {cap }}-I R-V=0} where\mathrm{V_{cap}} is the potential difference across capacitor

\mathrm{\therefore V_{\text {cap }} I R=\left(\frac{V}{3 R}\right) \times R=\frac{V}{3} }

Method II

Let us consider A to be at 0V. Then points B, C and D will be at V, V and 2V, respectively. Let the current be flowing in the clockwise direction. Applying Kirchhoff's law in the outer loop, we get \mathrm{V-I R-I(2 R)-2 V=0}

\mathrm{\therefore I=\frac{-V}{3 }R}. The minus sign here indicates that the current is in the opposite direction to what we have assumed. Applying Kirchhoff's law from A to E via B, we get \mathrm{ V_A+V+ I R=V_E}

\mathrm{ \therefore 0+V+\frac{V}{3 R} \times R=V_E=\frac{4 V}{3} }

Again applying Kirchhoff's law from A to E via C, we get

\mathrm{ V_A+V+V_{\text {cap }}=V_c \therefore V_{\text {cap }}=\frac{V}{3} }
 

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manish painkra

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