Question 3-17B5

To determine the impedance of a parallel R-L network, it's often easiest to first calculate the total admittance (Y), which is the reciprocal of impedance. 1. **Admittance of Components:** * Resistive Admittance ($G$) = $1/R = 1/300$ S * Inductive Susceptance ($B_L$) = $1/(jX_L) = 1/(j400) = -j(1/400)$ S (The negative 'j' indicates inductive reactance in admittance form). 2. **Total Admittance ($Y_{total}$):** * $Y_{total} = G + B_L = (1/300) - j(1/400)$ S $\approx 0.003333 - j0.0025$ S 3. **Convert Total Admittance to Polar Form:** * Magnitude $|Y_{total}| = \sqrt{(0.003333)^2 + (-0.0025)^2} \approx 0.004166$ S * Phase angle $\theta_Y = \arctan(-0.0025 / 0.003333) \approx -36.9^\circ$ 4. **Calculate Total Impedance ($Z_{total}$):** * $Z_{total} = 1/Y_{total}$ * Magnitude $|Z_{total}| = 1/|Y_{total}| = 1/0.004166 \approx 240$ ohms * Phase angle $\theta_Z = -\theta_Y = -(-36.9^\circ) = +36.9^\circ$ Therefore, the impedance is 240 ohms, /36.9 degrees. An inductive circuit, like this R-L parallel network, has a positive phase angle, meaning the voltage leads the current. Option A is incorrect because a negative phase angle implies a capacitive circuit. Options C and D are incorrect magnitudes and angles, likely resulting from treating the components as if they were in series.