Question
In the given circuit, with steady current, the potential drop across the capacitor must be
In the given circuit, with steady current, the potential drop across the capacitor must be
A.
$$V$$
B.
$$\frac{V}{2}$$
C.
$$\frac{V}{3}$$
D.
$$\frac{{2V}}{3}$$
Answer :
$$\frac{V}{3}$$
Solution :
There will be no current flowing in branch $$BE$$ in steady condition.
Let $$I$$ be the current flowing in the loop $$ABCDEFA.$$
Applying Kirchoff's law in the loop moving in anticlockwise direction starting from $$C.$$
$$\eqalign{ & + 2V - I\left( {2R} \right) - I\left( R \right) - V = 0 \cr & \therefore V = 3IR \cr & \Rightarrow I = \frac{V}{{3R}}\,\,......\left( 1 \right) \cr} $$
Applying Kirchoffs law in the circuit $$ABEFA$$ we get on moving in anticlockwise direction starting from $$B$$
$$\eqalign{ & + V + {V_{cap}} - IR - V = 0\left( {{\text{where}}\,{V_{cap}}\,{\text{is}}\,{\text{the}}\,p.d.\,{\text{across}}\,{\text{capacitor}}} \right) \cr & \therefore {V_{cap}} = IR = \left( {\frac{V}{{3R}}} \right) \times R = \frac{V}{3} \cr} $$
There will be no current flowing in branch $$BE$$ in steady condition.
Let $$I$$ be the current flowing in the loop $$ABCDEFA.$$
Applying Kirchoff's law in the loop moving in anticlockwise direction starting from $$C.$$
$$\eqalign{ & + 2V - I\left( {2R} \right) - I\left( R \right) - V = 0 \cr & \therefore V = 3IR \cr & \Rightarrow I = \frac{V}{{3R}}\,\,......\left( 1 \right) \cr} $$
Applying Kirchoffs law in the circuit $$ABEFA$$ we get on moving in anticlockwise direction starting from $$B$$
$$\eqalign{ & + V + {V_{cap}} - IR - V = 0\left( {{\text{where}}\,{V_{cap}}\,{\text{is}}\,{\text{the}}\,p.d.\,{\text{across}}\,{\text{capacitor}}} \right) \cr & \therefore {V_{cap}} = IR = \left( {\frac{V}{{3R}}} \right) \times R = \frac{V}{3} \cr} $$
