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

Options

A. V
B. `V//2`
C. `V//3`
D. `2V//3`

Answer

C

Explanation

c. Method I: There will be no current flowing in branch BE
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 C+
`2V - I(2R) -I(R) -V = 0 or V = 3IR or I = V//3R`.
Applying Kirchhoff's law in the
circuit `ABEF//A` we get on moving
in anticlockwise direciton starting
form B
`+V+V_(cap)-IR -V =0`
where `V_(cap)` is the potential
difference across capacitor
`V_(cap) = IR = V/(3R) xx R = V/3`
.
Method II:
let us consider A to be
at OV. Then points B,
C, and D will be at V, V,
and 2 V, respectively. Let
the current be flowing in
the clockwise direction.
Applying Kirchhoff's law
in the outer loop, we get
`V - IR - I(2R) -2V = 0 or I = -V//3R`.
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
`V_A + V + IR = V_E`
or `0+V+V/(3R) xx R = V_(E) = (4V)/3`
Again applying Kirchhoff's law from A to E via C, we get
`V_A + V + V_(cap) = V_(C) or V_(cap) = V/3`