In the simple AC circuit shown on the right, C = 0.051 F, L = 1.5 H, R = 32 ohm,

Question

In the simple AC circuit shown on the right, C = 0.051 F, L = 1.5 H, R = 32 ohm, delta V = delta V_max sin(omega t), where delta V_max = 52 V and omega = 95 rad/s. Part (a) Express the capacitive reactance, X_C, in terms of C and omega. Part (b) Calculate the value of X_C, in ohms. Part (c) Express the inductive reactance, X_L, in terms of L, and omega. Part (d) Calculate the value of X_L, in ohms. Part (e) Express the impedance, Z, in terms of R, X_l, and X_C. Z = 1/(RX_LX_C) Calculate the value of Z, in ohms. Part (g) Express the maximum current, I_max, in terms of delta V_max and Z. Part (h) Calculate the value of I_max, in amperes. Part (i) Calculate the value of rms current, I_rms, in amperes. Part (j) Express the average power delivered to the circuit, P_avg, in terms of I_rms and R. Part (k) Calculate the value of P_avg, in watts.

Explanation / Answer

part a )

Xc = 1/wc

part b )

given w = 95

c = 0.051

Xc = 0.206 ohm

XL = w*L

part d )

XL = 142.5 ohm

part e )

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

part f ) = 145.85 ohm

part g ) = Imax = Vmax/Z

part h ) = Imax = 52/145.85 = 0.36 A

part i ) Irms = Imax/sqrt(2)

Irms = 0.25 A

part j ) Pavg = Irms^2*R

part k )Pavg = 2.03 W

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