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Problem 22.18 A series ac circuit contains a 310 resistor, a 10.0 mH inductor, a

ID: 1543425 • Letter: P

Question

Problem 22.18

A series ac circuit contains a 310 resistor, a 10.0 mH inductor, a 3.10 F capacitor, and an ac power source of voltage amplitude 45.0 Voperating at an angular frequency of 360 rad/s .

Part A

What is the power factor of this circuit?

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Part B

Find the average power delivered to the entire circuit.

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Part C

What is the average power delivered to the resistor, to the capacitor, and to the inductor?

Problem 22.18

A series ac circuit contains a 310 resistor, a 10.0 mH inductor, a 3.10 F capacitor, and an ac power source of voltage amplitude 45.0 Voperating at an angular frequency of 360 rad/s .

Part A

What is the power factor of this circuit?

cos =

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Part B

Find the average power delivered to the entire circuit.

P=   W  

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Part C

What is the average power delivered to the resistor, to the capacitor, and to the inductor?

PR, PC, PL=   W

Explanation / Answer

RESISTANCE , R = 310 ohm

INDUCTIVE REACTANCE , XL = WL = 360*10*10^-3 = 3.6 ohm

CAPACITIVE REACTANCE , XC = 1/WC = 1/(360*3.10*10^-6) = 896.05 ohm

IMPEDENCE OF THE CIRCUIT

Z = [R^2 + (XC - XL)^2]^0.5 = [310^2 + (896.05 - 3.6)^2]^0.5 = 944.75

PART A) power factor

cos(theta) = R/Z = 310/944.75 = 0.328

PART B) since there is no average power delivered to inductor , capacitor

so average power deliverd to entire circuit = average power delivered to resistor

=> Pavg = V^2 / R = 45^2/310 = 6.532 W

PART C)

average power delivered to resistor, PR = 6.532 W

average power delivered to capacitor or inductor, PC = PL = 0 W

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