JEE Main 2026 — Physics Electromagnetism

In steady state, the capacitor acts as an open circuit, meaning no current flows through the branch containing the $100\ \mu\text{F}$100 μF capacitor. Consequently, no current flows through the $10\ \Omega$10 Ω resistor in series with it on the top right.

The voltage across the capacitor will be equal to the voltage across the $4\ \Omega$4 Ω vertical resistor.

Let's simplify the circuit to find this voltage. The $12\ \Omega$12 Ω resistor on the left is in parallel with the $12\ \text{V}$12 V battery, so the voltage across the rest of the circuit remains $12\ \text{V}$12 V.

For the rightmost part of the active circuit, the $4\ \Omega$4 Ω top resistor, the $4\ \Omega$4 Ω vertical resistor, and the $2\ \Omega$2 Ω bottom resistor are in series. Their equivalent resistance is:

$$R_{\text{right}} = 4\ \Omega + 4\ \Omega + 2\ \Omega = 10\ \Omega$$Rright​=4 Ω+4 Ω+2 Ω=10 Ω

This $10\ \Omega$10 Ω equivalent resistance is in parallel with the $10\ \Omega$10 Ω vertical resistor in the middle. The equivalent resistance of this parallel combination is:

$$R_{\text{mid}} = \dfrac{10 \times 10}{10 + 10} = 5\ \Omega$$Rmid​=10+1010×10​=5 Ω

Now, the total resistance of the main circuit connected to the $12\ \text{V}$12 V source is the sum of the $5\ \Omega$5 Ω top-left resistor, the $5\ \Omega$5 Ω equivalent middle resistance, and the $2\ \Omega$2 Ω bottom-left resistor:

$$R_{\text{total}} = 5\ \Omega + 5\ \Omega + 2\ \Omega = 12\ \Omega$$Rtotal​=5 Ω+5 Ω+2 Ω=12 Ω

The total current flowing from the $12\ \text{V}$12 V source into this main branch is:

$$I = \dfrac{12\ \text{V}}{12\ \Omega} = 1\ \text{A}$$I=12 Ω12 V​=1 A

The voltage across the parallel combination (the middle section) is:

$$V_{\text{mid}} = I \times R_{\text{mid}} = 1\ \text{A} \times 5\ \Omega = 5\ \text{V}$$Vmid​=I×Rmid​=1 A×5 Ω=5 V

This $5\ \text{V}$5 V is applied across the right branch which has a total resistance of $10\ \Omega$10 Ω. The current through this branch is:

$$I_{\text{right}} = \dfrac{5\ \text{V}}{10\ \Omega} = 0.5\ \text{A}$$Iright​=10 Ω5 V​=0.5 A

The voltage across the $4\ \Omega$4 Ω vertical resistor is:

$$V_{4\Omega} = I_{\text{right}} \times 4\ \Omega = 0.5\ \text{A} \times 4\ \Omega = 2\ \text{V}$$V4Ω​=Iright​×4 Ω=0.5 A×4 Ω=2 V

Since the capacitor is in parallel with this $4\ \Omega$4 Ω resistor (with no voltage drop across the $10\ \Omega$10 Ω resistor in its branch), the voltage across the capacitor is also $2\ \text{V}$2 V.

The stored charge in the capacitor is:

$$Q = C \times V = 100\ \mu\text{F} \times 2\ \text{V} = 200\ \mu\text{C}$$Q=C×V=100 μF×2 V=200 μC

Answer: $200$200