Skip to content

JEE MAIN 2026 · Current Electricity

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 49

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

The stored charge in the capacitor in steady state of the following circuit is ______ μC.

A 12 V battery drives a ladder network: 12 Ω directly across the battery; then 5 Ω in the top rail, 10 Ω as a rung, 4 Ω in the top rail, 4 Ω as a rung, and 10 Ω in the top rail leading to a 100 μF capacitor; the bottom rail has two 2 Ω resistors between the rungs

Your answer

Solution

Solution

In steady state no current flows into the capacitor, so the 10 Ω resistor in series with it carries no current and the capacitor sits across the 4 Ω rung.

Reduce the network. The 12 Ω resistor is straight across the battery and does not affect the rest. From the 5 Ω onwards:

  • The far loop (4 Ω top + 4 Ω rung + 2 Ω bottom) is 10 Ω10\ \Omega, in parallel with the 10 Ω rung: 5 Ω5\ \Omega.
  • Add the 5 Ω and the first 2 Ω in the bottom rail: 5+5+2=12 Ω5 + 5 + 2 = 12\ \Omega.

Current through this branch: 1212=1\frac{12}{12} = 1 A. Across the 10 Ω rung the voltage is 1×5=51 \times 5 = 5 V, so the far loop carries 510=0.5\frac{5}{10} = 0.5 A.

Capacitor voltage = voltage across the 4 Ω rung =0.5×4=2= 0.5 \times 4 = 2 V.

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

Answer: 200 μC.

Answer checked against the official answer key. The solution is our own working.

Keep going

The stored charge in the capacitor in… | SchoolToCareer