Question

A 1.15 μF capacitor, initially charged to 13.5 V , discharges when it is connected in series with a resistor.

Part A

What resistance is necessary to cause the capacitor to have only 37% of its initial charge 2.00 s after starting? (MΩ )

Part B

What is the voltage across the capacitor at t = 5 τ if the capacitor is instead charged by the same battery through the same resistor? (v)

Answer 1

Solution)

Given, q=1.15*20^-6 F

Q=CV= (1.15*10^-6)*13.5=1.55*10^-5 C

A) Q= Qo*e^-t/T

0.37 Qo= Qo*e^-t/T

0.37= exp(-t/T)

-t/T= ln(0.37)

T= -2secs/ ln(0.37)

T=2.011 secs

Now, T= RC

R= T/C= 2.011/1.15*10^-6 =1.7 Mohms (Ans)

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B)

V= E (1-e^-t/T)

V= 13.5( 1-e^-5)= 13.40 V (Ans)

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Answer 2

(A) Discharging of a capacitor is given by the following expression: V = Voe RC Here, V. is the initial voltage, V is the voltage after time t, R is the resistance, and C is the capacitance. Rearrange the above equation for R. R=- Cin Since V is 37 % less of V, hence, the above equation can be written as follows: R=- V-37 100 -Cin V -Cin 0.63 2.0 s -1.15 Fin 0.63 2.0 s (-1.15x10 F)(In 0.63) = 3.76x1062 =3.76 MO (B) Charging of capacitor is given by the following expression: - 13.50(1-27) = 13.41 V