Question

The capacitor in the circuit shown in (Figure 1) is charged to an initial value Q. When the switch is closed, it discharges through the resistor. It takes 1.5 seconds for the charge to drop to 12Q.

Part A

How long does it take to drop from initial charge Q to 14Q?

Express your answer to two significant figures and include the appropriate units.

Answer 1

let T is the time constant of the circuit.

we know, know charge on the capacitor at time t,

q = Qmax*e^(-t/T)

at t = 1.5 s

Q/2 = Q*e^(-1.8/T) --(1)

let time t, q = Q/4

Q/4 = Q*e^(-t/T)                        ---(2)

take equation(2) / equation(1)

2/4 = e^((-t/T)*(T/1.5)

0.5 = e^(-t/1.5)

ln(0.5) = -t/1.5

t = -1.5*ln(0.5)

   = 1.04 s ---> Answer

(or)

let T is the time constant of the circuit.

we know, know charge on the capacitor at time t,

q = Qmax*e^(-t/T)

at t = 1.5 s

12Q = Q*e^(-1.8/T) --(1)

let time t, q = 14Q

14Q = Q*e^(-t/T)                        ---(2)

take equation(2) / equation(1)

12/14 = e^((-t/T)*(T/1.5)

0.857 = e^(-t/1.5)

ln(0.857) = -t/1.5

t = -1.5*ln(0.857)

   = 0.231 s ---> Answer