Question

A 21.1 kΩ resistor and a capacitor are connected in series and then a 12.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 4.48 V in 1.32 µs. (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.

Answer 1

Solution)

Given,

Resistor, R=21.1*10^3 ohms

Voltage, Vo=12 V

Final Voltage, V=4.4 V

Time, t=1.32*10^-6 s

Part a)

We know,

V= Vo*(1-e^-t/RC)

Substitute values,

4.4=12*(1-e^(-1.32*10^-6/RC))

0.366=1-e^(-1.32*10^-6/RC)

After taking log on both sides,

-0.456=-1.32*10^-6/RC

So, time constant, RC= 2.89*10^-6 s=2.89 us(Ans)

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Part b)

Here, RC=2.89*10^-6 sec

So, Capacitance, C=(2.89*10^-6)/(21.1*10^3)= 1.37*10^-10 F (Ans)

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