Question

A circuit consists of an AC source, for which Emax = 18 V in series with a 1100-Ω resistor and a capacitor. The amplitude of the potential difference across the resistor is VR = 9.0 V when the source operates at 1700 Hz . What is the capacitance of the capacitor?

C =

Answer 1

Potential difference across resistor will be given by:

V_R = I_R*R

I_R = V_R/R

R = 1100 ohm & V_R = 9.0 V

I_R = 9.0/1100 = 8.18*10^-3 Amp

Now Since capacitor and resistor are both in series, So current in both resistor and capacitor will be same and equal to total current, So

Ic = 8.18*10^-3 Amp = I_R = I

So total impedance of circuit will be:

Emax = I*Z

Z = Emax/I = 18/(8.18*10^-3) = 2200.489

Now impedance of RC circuit will be:

Z = sqrt (R^2 + Xc^2)

Z^2 = R^2 + Xc^2

Xc = Capacitive reactance = 1/(w*C) = 1/(2*pi*f*C)

2200.489^2 = 1100^2 + 1/(2*pi*f*C)^2

1/(4*pi^2*f^2*C^2) = 2200.489^2 - 1100^2

f = 1700 Hz

So,

C = sqrt [1/(4*pi^2*1700^2*(2200.489^2 - 1100^2))]

C = 4.91*10^-8 F

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