(10%) Problem 7: In the simple AC circuit shown on the right, C-0.013 F, L-1.9 l

Question

(10%) Problem 7: In the simple AC circuit shown on the right, C-0.013 F, L-1.9 ll, R-39 , 1-AFmarsin(ot), where Almmx-61 Vand co-75 rads. AV ©theexpertta.cor 9% Part (a) Express the capacitive reactance, Xc, in terms of C and e. Grade Summary Deductions Potential 100 Sabmissions Attempts remaining 04% per attempt) detailed view 7 89 Subesit Hints:% dedaction per hint. Hints remaining Feedback: 5 deduction per feedback ii 9% Part (b) Calculate the value ofXC, in ohms 9% Part (c) Express the inductive reactance, Xi, in terms of L, and . 9% Part (d) Calculate the value of X., in ohms. 9% Part (e) Express the impedance, z, in terms of R, and Xc. 9% Part (f) Calculate the value of Z, in ohms 9% Part (g) Express the maximum current, inas, in terms of 'nn and Z. 9% Part (b) Calculate the value of Imax, in amperes. 9% Part (i) Calculate the value of rms current. /rms. In amperes. 9% Part(j) Express the average power delivered to the circuit. Pavp in terms of/m. and R. 9% Part (k) Calculate the value of Pave in watts.

Explanation / Answer

Given

L = 1.9 H , C = 0.013 F , R = 39 ohm

dV max = 61 V , W = 75 rad/s

LCR circuit

we know that the capacitive reactance is X_C = 1/W*C ohm

and X_C = 1/(75*0.013) ohm

X_C = 1.0256410 ohm

and the inductive reactance is XL = W*L

value is X_L = 75*1.9 ohm = 142.5 ohm

here X_L is > X_C so  

the impedance is Z = sqrt(R^2+(X_L-X_C)^2)

Z = sqrT(39^2+(142.5-1.0256410)^2) ohm

Z = 146.7515 ohm

maximum current will be at XL = XC

so Z = R

I_max = V/R = 61/39 = 1.5641 A

we have the relation between I_max and I_rms is I_rms = i_max/(sqrt(2))

I_rms = 1.5641/(sqrt(2))A = 1.106 A

AVerage power delivered is Pavg = Imax^2*R/2 = (sqrt(2) (I_rms))^2 *R/2

Pavg = (sqrt(2))(1.106))^2*39 /2

Pavg = 47.71 W

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