Question

A 30 μFμF capacitor initially charged to 30 μCμC is discharged through a 1.9 kΩkΩ resistor. How long does it take to reduce the capacitor's charge to 5.0 μCμC ?

Answer 1

Given Information:

Resistance = R = 1.9 k Ω

Capacitance = C = 30 uF

Initial charge = 30 uC

Final charge = 5 uC

Required Information:

Time taken to reduce the capacitor's charge to 5.0 μC = ?

Answer:

t = 0.101 seconds

Explanation:

The voltage across the capacitor is given by

V = V₀e^(–t/τ)

Where τ is time constant, V₀ is the initial voltage, V is the voltage after some time t

τ = RC

τ = 1900*30x10⁻⁶

τ = 0.057 sec

The initial voltage across the capacitor was

V₀ = Q/C

V₀ = 30/30

V₀ = 1 V

Voltage to reduce the charge to 5 uF

V = 5/30

V = 0.167 V

V = V₀e^(–t/τ)

0.167 = 1*e^(–t/0.057)

take ln on both sides

ln(0.167) = ln(e^(–t/0.057))

-1.789 = -t/0.057

t = 1.789*0.057

t = 0.101 seconds