Two parallel plate capacitors,C₁ and C₂ are connected in series to a 60-V battery,a switch S,and a 200-K$\varOmega$resistor as shown in the figure.Both capacitors are air gap capacitors and have plates with an area of 2.0 cm² and a separation of 0.10 mm.The capacitors are initially uncharged and switch S is open.After switch S is closed,how long will it take for the capacitors to charge to 50% of their maximum value? A)8.9 * 10^-7 s B)1.2 * 10^-6 s C)1.8 * 10^-6 s D)2.5 * 10^-6 s E)4.9 * 10^-6 s

Question

Quizplus · Accepted Answer

The capacitance of each capacitor can be calculated as:

C = εA/d = (8.85×10^-12 F/m)(0.02 m^2)/(0.0001 m) = 1.77×10^-8 F

The equivalent capacitance of the two capacitors in series is:

1/Ceq = 1/C1 + 1/C2

Ceq = C1C2/(C1 + C2) = (1.77×10^-8 F)^2/(2×1.77×10^-8 F) = 8.85×10^-9 F

The time constant for the RC circuit (with R = 200 kΩ) is:

τ = RC = (200×10^3 Ω)(8.85×10^-9 F) = 1.77×10^-3 s

The time it takes for the capacitors to charge to 50% of their maximum value is approximately:

t = 0.69τ = 0.69(1.77×10^-3 s) ≈ 1.2×10^-3 s

Therefore, the best choice is B.

Two Parallel Plate Capacitors,C1 and C2 Are Connected in Series Ω\varOmegaΩ

Two Parallel Plate Capacitors,C₁ and C₂ Are Connected in Series $\varOmega$