Capacitor Discharge Time

Electricity and Magnetism Level 2

A capacitor discharges through a resistor, which is connected across the capacitor terminals at t = 0 t = 0 . The capacitor is considered discharged if the charge it carries, Q C ≤ ϵ Q_C \le \epsilon for a certain small number ϵ \epsilon . Suppose it takes T T seconds to "discharge" the capacitor starting from a charge of Q 1 Q_1 , how much time (in seconds) will the capacitor need to discharge starting from an initial charge of 2 Q 1 2 Q_1 ?

Details and Assumptions:

  • The time constant of the capacitor-resistor circuit is 100 seconds.
T + 100 / ln ⁡ 2 T + 100 / \ln 2 T + 100 T + 100 T T T + 100 ln ⁡ 2 T + 100 \ln 2

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Hosam Hajjir
Mar 3, 2019

The discharge equation of the capacitor-resistor circuit is

Q ( t ) = Q ( 0 ) e − t T 0 Q(t) = Q(0) e^{- \dfrac{t}{T_0} }

where T 0 T_0 is the time constant, T 0 = 100 T_0 = 100 seconds. From the given information in the problem we can write,

ϵ = Q ( 0 ) e − T T 0 = 2 Q ( 0 ) e − T ′ T 0 \epsilon = Q(0) e^{- \dfrac{T}{T_0} } = 2 Q(0) e^{- \dfrac{T'}{T_0} }

Therefore,

T T 0 = − ln ⁡ 2 + T ′ T 0 \dfrac{T}{T_0} = - \ln 2 + \dfrac{T'}{T_0}

from which,

T ′ = T + T 0 ln ⁡ 2 = T + 100 ln ⁡ 2 T' = T + T_0 \ln 2 = T + 100 \ln 2

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