Answer:
19.4 ms
Step-by-step explanation:
The electric charge in a capacitor is given as;
Q= CV
Initial volts V = Q/C = 20µC/10µF = 2volts
final volts V = Q/C= 10µC/10µF = 1volts
Voltage on a capacitor undergoing discharge is given as;
v = v₀e^(–t/τ)
Where;
v₀ = initial voltage on the capacitor
v = voltage after time t
R = resistance in ohms,
C = capacitance in farads
t = time in seconds
RC = τ = time constant
Therefore;
τ = 10µF x 2.8kΩ = 28 ms
v = v₀e^(–t/τ)
1 = 2e^(–t/28ms)
e^(–t/28ms) = 0.5
–t/28ms = -0.693
t = (28ms)(0.693) = 19.4 ms
