In the steady-state circuit shown, branch currents are indicated. A 4 μF capacitor connects central nodes. The energy stored in the capacitor is x×10⁻⁴ J. Find x. (The current through the right-hand 2 Ω resistor is 1 A.)
Physics · Mixed Chapter
JEE Mains

Question

In the steady-state circuit shown, branch currents are indicated. A 4 μF capacitor connects central nodes. The energy stored in the capacitor is x×10⁻⁴ J. Find x. (The current through the right-hand 2 Ω resistor is 1 A.)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Explanation: In DC steady state the capacitor branch carries zero current. Apply KCL at the junctions and voltage drops I R along the resistors in the illustrated network. The resulting potential difference between capacitor plates is 20 V. Stored energy U=½CV²=½(4 μF)(20²)=8×10⁻⁴ J, giving x=8.