Let's break down and solve the problem step by step: **Given:** - Terminal voltage, \( V_{rms} = 250 \) V - Load absorbs average (real) power, \( P = 40 \) kW = \( 40,000 \) W - Load delivers magnetizing reactive power, \( Q = 30 \) kVAR = \( 30,000 \) VAR We are asked to **find the load resistance (\( R \)) and reactance (\( X \))**. --- ## **Step 1: Find Apparent Power and Load Impedance** The complex power \( S \) is: \[ S = P + jQ \] \[ |S| = \sqrt{P^2 + Q^2} \] Plug in the values: \[ |S| = \sqrt{(40,000)^2 + (30,000)^2} \] \[ = \sqrt{1,600,000,000 + 900,000,000} \] \[ = \sqrt{2,500,000,000} \] \[ = 50,000 \text{ VA} \] --- ## **Step 2: Find the Load Current** \[ |S| = V_{rms} \cdot I_{rms} \] So, \[ I_{rms} = \frac{|S|}{V_{rms}} \] \[ I_{rms} = \frac{50,000}{250} = 200 \text{ A} \] --- ## **Step 3: Find Load Impedance** The impedance \( Z \) is: \[ Z = \frac{V_{rms}}{I_{rms}} \] \[ Z = \frac{250}{200} = 1.25 \ \Omega \] --- ## **Step 4: Find Power Factor Angle** The power factor angle \( \theta \) is: \[ \cos\theta = \frac{P}{|S|} \] \[ \cos\theta = \frac{40,000}{50,000} = 0.8 \] Since the load is **delivering a magnetizing reactive power**, that means **inductive** (positive Q). So: \[ \tan\theta = \frac{Q}{P} = \frac{30,000}{40,000} = 0.75 \] \[ \theta = \arccos(0.8) \approx 36.87^\circ \] --- ## **Step 5: Find Resistance and Reactance** The impedance in polar form: \[ Z = |Z| \angle \theta = 1.25 \angle 36.87^\circ \] In rectangular form: \[ Z = R + jX \] \[ R = |Z| \cos\theta = 1.25 \times 0.8 = 1.0\ \Omega \] \[ X = |Z| \sin\theta = 1.25 \times 0.6 = 0.75\ \Omega \] --- ## **Step 6: Final Answers** \[ \boxed{ \begin{aligned} \text{Load Resistance, } R &= 1.0\ \Omega \\ \text{Load Reactance, } X &= 0.75\ \Omega \\ \end{aligned} } \] --- **Summary Table:** | Quantity | Value | |-----------------|--------------| | Resistance (R) | \(1.0~\Omega\) | | Reactance (X) | \(0.75~\Omega\) | --- **Explanation:** - The load absorbs 40 kW (real power) and 30 kVAR (inductive reactive power). - Apparent power is 50 kVA, so current is 200 A. - Impedance is 1.25 Ω at 36.87° (power factor angle). - Convert to rectangular components: \( R = 1.0~\Omega \), \( X = 0.75~\Omega \).