Question

In the circuit shown the capacitor and an ideal inductor do not have energy initially and key is switch on at $t=0$. Then which of the following options is/are true:

• A. Potential difference across the capacitor at $t=0$ is zero
• B. The current flowing through the battery at $t=0$ is $0.5A$
• C. Potential difference across the inductor at $t=\infty$ is zero
• D. The current flowing through the battery at $t=\infty$ is $\frac{6}{11}$ A

Answer 1

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

The correct options are
A Potential difference across the capacitor at $t=0$ is zero
B The current flowing through the battery at $t=0$ is $0.5A$
C Potential difference across the inductor at $t=\infty$ is zero
D The current flowing through the battery at $t=\infty$ is $\frac{6}{11}$ A

key : As we know that at t = 0 capacitor offers zero resistance and at t = infinity capacitors get fully charged and offers infinite resistance (in another term we assume break in circuit ).
While inductor offers infinite resistance at t = 0 and zero resistance at t = infinite (in fully charged condition).
So According to above concept we draw the circuit diagram at t = 0 and t = infinity as shown.
so at t = 0
Resistance of capacitor = 0, so potential difference = 0
resistance of 10$\Omega$ is out of circuit because current choose to follow the path having less resistance and capacitor has zero resistance at t=0. So current will follow this path only.
So according to image (i), at t = 0
Total resistance of circuit (R) = $20+20=40$ $\Omega$
Total potential difference (V) = $20V$
Current I = $\frac{V}{R}$ ,
Therefore I= $\frac{20}{40}=0.5A$
At t = $\infty$
Resistance due to inductor is zero so potential difference across inductor at t = $\infty$ is zero.
according to image (ii) solving circuit.
Last two 10$\Omega$ and 20$\Omega$ resistances are in parallel, so its resultant resistance (R1) = $\frac{20\times 10}{20+10}=\frac{20}{3}\Omega$
First two resistances 10, 20 and resultant R1 are in series.
So total resistance of circuit ($R_2$) = $10+20+\frac{20}{3}=\frac{110}{3}\Omega$
Current at t = $\infty$ , I = $\frac{V}{R2}$
I = $\frac{20}{\frac{110}{3}}=\frac{6}{11}A.$
All options are true.