Question

A capacitor of2 F (practically not possible to have a capacity of 2 F) is charged by a battery of 6 V, The battery is removed and circuit is made as shown, Switch is closed at time $t=0$. Choose the correct options

• A. at time $t=0$ current in the circuit is 2 A
• B. at time $t=\left(6ln2\right)$ second, potential difference across capacitor is 3 V
• C. at time $t=\left(6ln2\right)$ second, potential difference across $1\Omega$ resistance is 1 V
• D. at time $t=\left(6ln2\right)$ second, potential difference across $2\Omega$ resistance is 2 V

Answer 1

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

The correct options are
A at time $t=0$ current in the circuit is 2 A
B at time $t=\left(6ln2\right)$ second, potential difference across capacitor is 3 V
C at time $t=\left(6ln2\right)$ second, potential difference across $1\Omega$ resistance is 1 V
D at time $t=\left(6ln2\right)$ second, potential difference across $2\Omega$ resistance is 2 V

At $t=0$ the emf of circuit $=$ potential difference across capacitor $=6V$
Thus, current in the circuit is $I=\frac{6}{1+2}=2V$
During discharging, the potential across capacitor at time t is
$V\left(t\right)=V_0{e}^{(-t/RC)}$ here time constant $=RC=(1+2)2=6s$
at $t=6ln2,V=6e^{(}-6ln2/6)=6e^{l}n{2}^{-1}=6(1/2)=3V$
Now current through the circuit is $I^{\prime }=V/(1+2)=3/3=1V$
Potential across resistor $1\Omega =1\times 1=1V$ and potential across resistor $2\Omega =2\times 1=2V$.