Question

An L-R-C series circuit consists of a 2.40 μF capacitor, a 6.00 mH inductor, and a 60.0 Ω resistor connected across an ac source of voltage amplitude 10.0 V having variable frequency.

Part A: At what frequency is the average power delivered to the circuit equal to 1/2*V_rms*I_rms? (ω = ____rad/s)

Part B: Under the conditions of part (a), what is the average power delivered to each circuit element? (P_R, P_C, P_L = ____W)

Part C: What is the maximum current through the capacitor? (I_C = ____A)

Answer 1

Part A.

Average power in LRC circuit is given by:

Pavg = Vrms*Irms*cos

Now given that average power is equal to

Pavg = (1/2)Vrms*Irms

So

Pavg = Vrms*Irms*cos = (1/2)*Vrms*Irms

cos = 1/2

= phase angle = arccos (1/2) = 60 deg

Now phase angle is given by;

= arctan ((XL - Xc)/R)

XL = w*L

Xc = 1/(w*C)

So using given values:

tan = tan 60 deg = sqrt 3

sqrt 3 = (w*6.00*10^-3 - 1/(w*2.40*10^-6))/60.0

Solving above quadratic equation using scientific calculator:

w = 20678.8 rad/sec

Part C.

Imax = Vmax/Z

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

Z = sqrt (60^2 + (20678.8*6.00*10^-3 - 1/(20678.8*2.40*10^-6)^2))

Z = 120.0 Ohm

So

Imax = Vmax/Z = 10.0/120.0 = 0.0833 Amp

Part B.

Now Average power in capacitor and Inductor will always be zero, So

P_R = Vrms*Irms*cos

Vrms = Vmax/sqrt 2

Irms = Imax/sqrt 2

So,

P_R = (Vmax*Imax*cos )/2

P_R = (10.0*0.0833*cos 60 deg)/2

P_R = 0.21 W

P_C = 0 W

P_L = 0 W

P_R, P_C, P_L = 0.21, 0, 0 W