In a series RL circuit, E_T = 120 V, R = 500 Ω , and X_L = 500 Ω . What is the true power? A) 8.8 W B) 14.4 W C) 17.2 W D) 20.36 W

Question

Quizplus · Accepted Answer

True power in an RL circuit is calculated using the formula $P = \frac{V^2}{Z^2} \times R$, where $V$ is the voltage across the circuit, $R$ is the resistance, and $Z$ is the impedance of the circuit. The impedance $Z$ of an RL circuit is found by $Z = \sqrt{R^2 + X_L^2}$, where $X_L$ is the inductive reactance. Given $R = 500 \Omega$ and $X_L = 500 \Omega$, the impedance $Z = \sqrt{500^2 + 500^2} = \sqrt{250000 + 250000} = \sqrt{500000} = 707.1 \Omega$. The true power $P = \frac{120^2}{707.1^2} \times 500 = \frac{14400}{500000} \times 500 = 14.4 W$.

In a Series RL Circuit, ET = 120 V, R

In a Series RL Circuit, E_T = 120 V, R