In the circuit shown, the energy stored in the capacitor is n μ J. The value of n is __________

Solution

Physics sir · 4 steps

In steady state the capacitor blocks current, so find the voltage across it from the two resistor branches.

1. Steady state. Once the capacitor is fully charged, no current flows through it. The circuit becomes two separate branches across the 12 V battery: 3 Ω + 9 Ω on the left and 4 Ω + 2 Ω on the right.
2. Left junction. The left branch has 12 Ω, so its current is 12/12 = 1 A. Take the bottom wire as 0 V and the top as 12 V. The 3 Ω resistor drops 3 V, so the left plate is at 12 - 3 = 9 V.
3. Right junction. The right branch has 6 Ω, so its current is 12/6 = 2 A. The 4 Ω resistor drops 8 V, so the right plate is at 12 - 8 = 4 V. The capacitor therefore sees V = 9 - 4 = 5 V.

   U = 1/2CV² = 1/2 × 6 μF × (5 V)² = 75 μJ

4. So. The energy stored is 75 μJ, so n = 75.

The common slip: 150. Leaving out the 1/2 in 1/2CV² gives 6 × 25 = 150, double the right value.

The idea to keep. A charged capacitor in a DC circuit is an open switch. Find the potential at each of its plates using the resistors alone, then use 1/2CV² with their difference.

Answer from the NTA final answer key. Solution by PrepWiser.