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

Question

In the given circuit, with steady current, the potential drop across the capacitor must be (A) V (B) V/2 (C) V/3 (D) 2V/3

Tardigrade Admin · Accepted Answer

Moving anticlockwise from A, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/057373ceeaf675c0d502a48cdbadaaad-.png" /> -i R-V+2 V-2 i R=0 or 3 i R=V or i=(V/3 R) VA-VB=i R+V-V=i R ⇒ Potential drop across C=(V/3)

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

V

V/2

V/3

2V/3

Moving anticlockwise from $A$ ,

$- i R - V + 2 V - 2 i R = 0$
or $3 i R = V$ or $i = \frac{V}{3 R}$
$V_{A} - V_{B} = i R + V - V = i R$
$\Rightarrow$ Potential drop across $C = \frac{V}{3}$

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

V

V/2

V/3

2V/3

Moving anticlockwise from A, −iR−V+2V−2iR=0 or 3iR=V or i=3RV​ VA​−VB​=iR+V−V=iR ⇒ Potential drop across C=3V​

Moving anticlockwise from $A$ ,

$- i R - V + 2 V - 2 i R = 0$
or $3 i R = V$ or $i = \frac{V}{3 R}$
$V_{A} - V_{B} = i R + V - V = i R$
$\Rightarrow$ Potential drop across $C = \frac{V}{3}$