In the shown circuit, the energy stored in the 6 μF capacitor is n μJ. Find n.

Question

G1 Physics · Accepted Answer

From the circuit: Current in the 3 Ω + 9 Ω branch: I₁ = 12/(3 + 9) = 1 A Current in the 4 Ω + 2 Ω branch: I₂ = 12/(4 + 2) = 2 A Potential drop from A to C: Vₐ - V꜀ = 3I₁ = 3 V Potential drop from A to D: Vₐ - Vᴅ = 4I₂ = 8 V Therefore capacitor voltage magnitude is V꜀ᴅ = 8 - 3 = 5 V Energy stored: U = ½CV² = ½ × 6 μF × 5² = 75 μJ Thus n = 75. Final answer: 75.