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Mice21 [21]
3 years ago
6

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 disconnect

ed 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)?
Physics
1 answer:
Iteru [2.4K]3 years ago
5 0

Answer:

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

V = V_{o} e^{\frac{-t}{T}}

2.2 = 6 e^{\frac{-t}{66}}

t = 66.2 sec

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