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Please help me with the following problem. I will rate LIFESAVER!Thanks for the

ID: 1674949 • Letter: P

Question

Please help me with the following problem. I will rate LIFESAVER!Thanks for the help.

Consider a 18, a 30F capacitor, and a 50mHinductor.

A) What is the peak current when just the inductor is connectedacross a standard outlet?

The answer should be 9.00A. Can you show me how to get this?

B) What is the rms current when just the resistor and capacitor areconnected in series across an AC source with an rms voltage of 4Vand a frequency of 100 Hz?

The answer is 71.4 mA. Can you show me how to get this?

C) What is the average power dissipated in the circuit in partb?

D) Suppose the capacitor is given a charge of 5 C andconnected across the resistor. How long does it take for the chargeto reach 1 C? (Only the capacitor and resistor are connectedhere)

Thanks for the help! Will rate LIFESAVER!

Explanation / Answer

Part A: Standard wall supply is 120 V and 60 HZ V peak = (2)*120 = 169.71 V Inductive Reactance = 2*60*50*10^-3 = 18.85 Peak current = Ipeak = 169.71/18.85 = 9.0031 = 9 A Part B: V rms = 4 V f = 100 HZ R = 18 C = 30 F = 3*10^-5 Reactance of the circuit = (R^2 + 1/(C)^2 R^2 = 18*18 = 324 Xc = 1/2fC = 1/(2*100*3*10^-5) = 53.05 Xc*Xc = 53.05^2 = 2814.5 Circuit Reactance = (324+2814.5) = 56.022 Irms = 4/56.022 = 71.4003 mA = 71.4 mA Part C: Avearge power = Vrms*Irms*cos tan = Xc/R = 53.05/18 = 2.97 = 71.4 cos = 0.321 Average power = 4*71.4*10^-3*0.321 = 606.6 milli watts Part D: q = 5 C 1 C = 5 C*e-t/RC t= 144.8S Part B: V rms = 4 V f = 100 HZ R = 18 C = 30 F = 3*10^-5 Reactance of the circuit = (R^2 + 1/(C)^2 R^2 = 18*18 = 324 Xc = 1/2fC = 1/(2*100*3*10^-5) = 53.05 Xc*Xc = 53.05^2 = 2814.5 Circuit Reactance = (324+2814.5) = 56.022 Irms = 4/56.022 = 71.4003 mA = 71.4 mA Part C: Avearge power = Vrms*Irms*cos tan = Xc/R = 53.05/18 = 2.97 = 71.4 cos = 0.321 Average power = 4*71.4*10^-3*0.321 = 606.6 milli watts Part D: q = 5 C 1 C = 5 C*e-t/RC t= 144.8S