JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

The charge stored by the capacitor C in the given circuit in the steady state is …….. $μ$ C.

A
12.5
B
10
C
7.5
D
5

Solution

In steady state, no current flows through the capacitor branch. Of the two diode branches, only the forward-biased one conducts; the reverse-biased branch is open. The active loop is therefore the 2.5 V source through the 1 $Ω$ series resistor and the 4 $Ω$ branch (the conducting diode path), giving a total of 5 $Ω$ .

Loop current: $i = \frac{2.5}{5} = 0.5$ A.

The capacitor sits in parallel with the 4 $Ω$ resistor in the conducting path, so $V_{C} = 0.5 \times 4 = 2$ V.

Charge: $Q = C V_{C} = 5 \times 2 = 10$ $μ$ C.

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