JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Capacitance

In the circuit shown, initially there is no charge on capacitors and keys ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ are open. The values of the capacitors are ${\mathrm{C}}_1=10\mu \mathrm{F},{\mathrm{C}}_2=30\mu \mathrm{F}$ and ${\mathrm{C}}_3={\mathrm{C}}_4=80\mu \mathrm{F}$.

Which of the statement(s) is/are correct ?

1. (A)The keys ${\mathrm{S}}_1$ is kept closed for long time such that capacitors are fully charged. Now key ${\mathrm{S}}_2$ is closed, at this time, the instantaneous current across $30\Omega$ resistor (between points P and Q) will be 0.2 A (round off to $1^{\text{st}}$ decimal place).
2. (B)If key ${\mathrm{S}}_1$ is kept closed for long time such that capacitors are fully charged, the voltage difference between points P and Q will be 10 V.
3. (C)At time $\mathrm{t}=0$, the key ${\mathrm{S}}_1$ is closed, the instantaneous current in the closed circuit will be 25 mA.✓ Correct
4. (D)If key ${\mathrm{S}}_1$ is kept closed for long time such that capacitors are fully charged, the voltage across the capacitors ${\mathrm{C}}_1$ will be 4V.✓ Correct

Answer: (C), (D)

Solution

**At t=0 when $S_1$ is closed:** Capacitors act as short circuits.

$$i=\frac{V}{R_{eq}}=\frac{5}{200}=25\text{ mA}$$

Option (C) is correct.

At steady state when capacitors are fully charged: They act as open circuits. The capacitors are in series, so charge on each is the same.

$$Q=C_{eq}V=8\mu F\times 5V=40\mu C$$

Voltage across $C_1$:

$$V_{C_1}=\frac{Q}{C_1}=\frac{40\mu C}{10\mu F}=4V$$

Option (D) is correct.

Voltage between P and Q:

$$V_P-\frac{40}{10}=V_Q$$

$$V_P-V_Q=4V$$

Option (B) is incorrect (states 10V).

For option (A), when $S_2$ is closed after steady state, the analysis shows the current is not 0.2A, so option (A) is incorrect.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options