Question

A capacitor of $2 F$ (theoretical value) capacitance is charged by a battery of $6\, V$. The charging battery is disconnected and circuit is made as shown. If the switch is closed at $t=0$. Which of the following options is/are correct

• A. At time $t=0$ current in the circuit is $2 \, A$
• B. At time $t=6 \ln (2)$ second potential difference across capacitor is $3 \, V$
• C. At time $t=6 \ln (2)$ second, potential difference across $1 \, \Omega$ resistance is $1 \, V$
• D. At time $t=6 \ln (2)$ second potential difference across $2 \, \Omega$ resistance is $2\, V$

Answer 1

Answer: (A), (B), (C), (D)

At $t=0$ just after closing the switch capacitor behaves like EMF $6\, V$ and circuit current will be
$i=\frac{6}{1+2}=2\, A$
Half life time of the circuit can be given as
$t_{1 / 2}=(\ln 2) \tau_{C}=R C(\ln 2)=6 \ln (2) s$
In half-life time all circuit parameters get halved Potential difference across capacitor will be
$V_{C}=\frac{6}{2}=3 \,V$
Current in circuit will be
$i=\frac{2}{1}=1\, A$
Potential differences across $1 \,\Omega$ and $2\, \Omega$ resistances will be
$V_{1 \Omega}=i R=1 \,V $
$V_{2 \Omega}=i R=2 \,V$