Capacitors, c1 = 1.0 f and c2 = 1.0 f, are connected in parallel to a 6.0 volt battery (δv = 6.0v). if the battery is disconnected and the capacitors are connected to a 33 ohm resistor, how long should it take for the voltage to cross the capacitors to drop to 2.2 volts (36.8% of the original 6.0 volts)?

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

C = 1.0 + 1.0

C = 2.0 F

R = resistance = 33 Ω

Time constant is given as

T = RC

T = 33 x 2

T = 66 sec

V₀ = initial potential difference across the combination = 6.0 Volts

V = final potential difference = 2.2 volts

Using the equation

t = 66.2 sec