Asked by Pranshu Saini on May 16, 2024
Verified
In a series RL circuit, ET = 120 V, R = 500 Ω , and XL = 500 Ω . What is the true power?
A) 8.8 W
B) 14.4 W
C) 17.2 W
D) 20.36 W
True Power
The actual power consumed by a circuit, calculated as the product of the voltage, current, and the cosine of the phase angle between them in a resistive load, measured in watts.
Series RL Circuit
An electrical circuit composed of resistors (R) and inductors (L) connected in a series configuration, affecting the phase and amplitude of the circuit's response.
Inductive Reactance
The resistance offered by an inductor to alternating current, which varies with the frequency of the current and the inductance, also measured in ohms.
- Describe and calculate true power in series RL circuits and its relationship with other power types.
Verified Answer
ZK
Zybrea KnightMay 18, 2024
Final Answer :
B
Explanation :
True power in an RL circuit is calculated using the formula P=V2Z2×RP = \frac{V^2}{Z^2} \times RP=Z2V2×R , where VVV is the voltage across the circuit, RRR is the resistance, and ZZZ is the impedance of the circuit. The impedance ZZZ of an RL circuit is found by Z=R2+XL2Z = \sqrt{R^2 + X_L^2}Z=R2+XL2 , where XLX_LXL is the inductive reactance. Given R=500ΩR = 500 \OmegaR=500Ω and XL=500ΩX_L = 500 \OmegaXL=500Ω , the impedance Z=5002+5002=250000+250000=500000=707.1ΩZ = \sqrt{500^2 + 500^2} = \sqrt{250000 + 250000} = \sqrt{500000} = 707.1 \OmegaZ=5002+5002=250000+250000=500000=707.1Ω . The true power P=1202707.12×500=14400500000×500=14.4WP = \frac{120^2}{707.1^2} \times 500 = \frac{14400}{500000} \times 500 = 14.4 WP=707.121202×500=50000014400×500=14.4W .
Learning Objectives
- Describe and calculate true power in series RL circuits and its relationship with other power types.