Question: Find $I_L$ and the circuit pf. 208 V motor $R = 20 \ \Omega$ pf = .874 $x_c = 50 \ \Omega$ top-left: a horizontal 208 V supply with a motor branch labeled $R = 20 \ \Omega$ and pf = .874 in the center, and a capacitor branch labeled $x_c = 50 \ \Omega$ on the right. g. Find $I_L$ and the circuit pf 208 V $R = 30 \ \Omega$ $R = 5 \ \Omega$ $X_L = 80 \ \Omega$ bottom-left: a horizontal 208 V supply with a resistor branch labeled $R = 30 \ \Omega$ in the center and an inductor branch labeled $R = 5 \ \Omega$, $X_L = 80 \ \Omega$ on the right.

1. **Problem Statement:** Find the load current $I_L$ and the power factor (pf) of the circuit given the following data: - Supply voltage $V = 208$ V - Motor branch resistance $R = 20 \ \Omega$ with power factor $pf = 0.874$ - Capacitive reactance $x_c = 50 \ \Omega$ 2. **Step 1: Calculate the motor current $I_m$** The power factor $pf$ is the cosine of the phase angle $\theta$ between voltage and current: $$pf = \cos \theta = 0.874$$ Calculate $\theta$: $$\theta = \cos^{-1}(0.874) \approx 29.1^\circ$$ The motor impedance $Z_m$ magnitude is: $$Z_m = R = 20 \ \Omega$$ Since only resistance is given, the motor current magnitude is: $$I_m = \frac{V}{Z_m} = \frac{208}{20} = 10.4 \ \text{A}$$ 3. **Step 2: Calculate the reactive component of motor current $I_{mQ}$** Reactive current component: $$I_{mQ} = I_m \sin \theta = 10.4 \times \sin 29.1^\circ \approx 10.4 \times 0.486 = 5.06 \ \text{A}$$ Active current component: $$I_{mP} = I_m \cos \theta = 10.4 \times 0.874 = 9.09 \ \text{A}$$ 4. **Step 3: Calculate the capacitive current $I_c$** Capacitive reactance $x_c = 50 \ \Omega$ means capacitive current leads voltage by 90 degrees: $$I_c = \frac{V}{x_c} = \frac{208}{50} = 4.16 \ \text{A}$$ This current is purely reactive and leads voltage, so it subtracts from the motor reactive current. 5. **Step 4: Calculate total reactive current $I_Q$** $$I_Q = I_{mQ} - I_c = 5.06 - 4.16 = 0.9 \ \text{A}$$ 6. **Step 5: Calculate total current $I_L$** Total current magnitude is: $$I_L = \sqrt{I_{mP}^2 + I_Q^2} = \sqrt{9.09^2 + 0.9^2} = \sqrt{82.64 + 0.81} = \sqrt{83.45} \approx 9.13 \ \text{A}$$ 7. **Step 6: Calculate circuit power factor $pf_{circuit}$** $$pf_{circuit} = \frac{I_{mP}}{I_L} = \frac{9.09}{9.13} \approx 0.996$$ This means the circuit power factor is approximately 0.996 (almost unity). --- **Final answers:** $$I_L \approx 9.13 \ \text{A}$$ $$pf_{circuit} \approx 0.996$$