Question Detail

PHY 102 Test Compilation PHY 102 16 Objective Question

Question

A given RLC circuit has R=100Ω, L=1H and C=10µF all connected in series with a source of 200sinωt with ω = 400 rad/s. Calculate the phase angle and determine the circuit type.
Options
A 161.5°, 86.5°
B 56.3°, capacitive
C 475.4°, 86.5°
D 56.3°, inductive
Correct Answer
Correct Answer

Option D is the correct answer.

Detailed Explanation

First, calculate the inductive reactance (XL) and capacitive reactance (XC):Given:Resistance R = 100 ΩInductance L = 1 HCapacitance C = 10 µF = 10 × 10-6 F = 10-5 FAngular frequency ω = 400 rad/sXL = ωL = 400 rad/s × 1 H = 400 ΩXC = 1/(ωC) = 1/(400 rad/s × 10-5 F) = 1/(4 × 10-3) = 1000/4 = 250 ΩNext, determine the net reactance X = XL - XC:X = 400 Ω - 250 Ω = 150 ΩSince XL > XC, the circuit is inductive.Finally, calculate the phase angle φ = arctan(X/R):φ = arctan(150 Ω / 100 Ω) = arctan(1.5)φ ≈ 56.3°Therefore, the phase angle is 56.3° and the circuit type is inductive.

Hint

Calculate inductive reactance (XL) and capacitive reactance (XC) first. The sign of (XL - XC) determines if the circuit is inductive or capacitive. Then use the formula φ = arctan((XL - XC)/R) for the phase angle.

Question Info