Question Detail

PHY 102 Test Compilation PHY 102 16 Objective Question
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Question

An inductor, L = 0.5H, in series with a resistor, R = 250Ω, are connected to an a.c. source 100V, 50Hz. Determine the apparent power of the circuit.
Options
A 33.9VA
Correct Answer
B 34VA
C None
D None
Correct Answer

Option A is the correct answer.

Detailed Explanation

Apparent power (S) in an AC circuit is given by S = Vrms * Irms or S = Vrms2 / Z, where Z is the impedance.Given: Vrms = 100 V, R = 250 Ω, L = 0.5 H, f = 50 Hz.First, calculate the angular frequency ω = 2πf = 2π(50 Hz) = 100π rad/s.Next, calculate the inductive reactance XL = ωL = (100π rad/s) * (0.5 H) = 50π Ω ≈ 157.08 Ω.Now, calculate the impedance Z = √(R2 + XL2) = √(2502 + (50π)2) = √(62500 + 24674.01) = √87174.01 ≈ 295.25 Ω.Finally, calculate the apparent power S = Vrms2 / Z = (100 V)2 / 295.25 Ω = 10000 / 295.25 ≈ 33.87 VA.Rounding to one decimal place, S ≈ 33.9 VA.

Hint

Apparent power is Vrms2/Z. Remember to calculate the inductive reactance and then the total impedance before finding the apparent power.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 16
Type: Objective
Difficulty: Not specified
Question 1083 of 1819
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