Question

The 2-microfarad (2 × 10^–6 farad) capacitor shown in the circuit above is fully charged by closing switch S₁ and keeping switch S₂  open,thus connecting the capacitor to the 2,000-volt power supply.

a. Determine each of the following for this fully charged capacitor.

i. The magnitude of the charge on each plate of the capacitor.

ii. The electrical energy stored in the capacitor.

At a later time, switch S₁ is opened. Switch S₂ is then closed, connecting the charged 2-microfarad capacitor to a 1-megohm (1 × 10⁶Ω) resistor and a 6-microfarad capacitor, which is initially uncharged.

b. Determine the initial current in the resistor the instant after switch S₂ is closed.

Equilibrium is reached after a long period of time.

c. Determine the charge on the positive plate of each of the capacitors at equilibrium.

d. Determine the total electrical energy stored in the two capacitors at equilibrium. If the energy is greater than the energy determined in part a. ii., where did the increase come from? If the energy is less than the energy determined in part a. ii., where did the electrical energy go?

Answer 1

a.

b. When the switch is closed, there is no charge on the 6 μF capacitor so the potential difference across the resistor equals that across the 2 μF capacitor, or 2000 V and I = V/R = 2 × 10^-3 A

c. In equilibrium, charge is no longer moving so there is no potential difference across the resistor therefore the capacitors have the same potential difference. V₂ = V₆ gives Q₂/C₂ = Q₆/C₆ giving Q₆ = 3Q₂ and since total charge is conserved we have Q₂ + Q₆ = Q₂ + 3Q₂ = 4Q₂ = 4 × 10^–3 C so Q₂ = 1 × 10^–3 C and Q₆ = 3 × 10^–3 C

d. U_C = U₂ + U₆ = Q₂²/2C₂ + Q₆²/2C₆ = 1 J. This is less than in part a. ii. Part of the energy was converted to heat in the resistor.