# RLC values in an AC circuit

A series RLC circuit has R=150 ohm, L=500 mH, C=30 uF is connected to a signal generator, whose output frequency is adjusted to 100 Hz and the averaged output voltage is 10.0V (a) Calculate the impedance Z of the circuit; (b) Find the voltage on each element measured by a voltmeter. (c) Calculate the power factor cos(theta). (d) If the value of the capacitor is suspected to be lower than the stated value, find a way to measure the exact value of the capacitor using the equipment (the signal generator and digital multi-meters) provided.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

a.

XL = 2*pi*f*L = 2*pi*100*0.5 = 314.16 ohm

XC = 1/(2*pi*f*C) = 1/(2*pi*100*30*10^-6) = 53.05 ohm

Impedance, Z = sqrt (R^2 + (XL -XC)^2) = sqrt(150^2 + (314.16 - 53.05)^2) = 301.13 ohm

b.

I = V/Z = 10 / 301.13 = 0.033 Amp

VL = I * XL = 10.43 Volt

VC = I*XC = 1.76 Volt

VR = I*R = 4.98 Volt

c.

Power factor,

cos(phi) = R/Z = 150/301.13 = 0.498

d.

If C is less than the stated value, XC will be greater than the estimated value, and in-turn impedance will less, and current will be more. Hence, measure voltage (using multimeter) across capacitor will be greater than the estimated value.

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