Question

Find the potential drop across $4\mu F$ capacitor and $6\Omega$ resistor in figure.

• A. 0, 0
• B. 0, 3V
• C. 0, 2V
• D. 2V, 0

Answer 1

The correct option is D 2V, 0

A fully charged capacitor draws no current, hence no current flows through the branch having capacitor.
The current through the circuit is
$I=\frac{V}{R}$
where, R is the total resistance of the circuit except in the branch having capacitor.
$I=\frac{3}{2+4}$
$I=\frac{3}{6}=\frac{1}{2}A$
Now, the potential drop across $4\Omega$ resistor is
$V=\frac{I}{R}$
$V=\frac{1}{2}\left(4\right)$
$V=2V$
Also, the potential drop across $6\Omega$ resistor is
$V=\frac{I}{R}$
Since, no current flows through $6\Omega$ resistor, hence
$V=\frac{0}{6}$
$V=0V$
The potential drop across $4\mu F$ capacitor is equal to the potential drop across $4\Omega$ resistor, since no current flows through the branch i.e. the potential drop across $6\Omega$ resistor is zero.