Question E5D11:
How much real power is consumed in a circuit consisting of a 100-ohm resistor in series with a 100-ohm inductive reactance drawing 1 ampere?  

Choices:  
A: 70.7 watts  
B: 100 watts  
C: 141.4 watts  
D: 200 watts    

Correct Answer is  B: 100 watts  

To find real power in a series R–L circuit, remember only the resistor dissipates real power.

Given: $R = 100\,\Omega$, $X_L = 100\,\Omega$, $I = 1\,\text{A}$.

Real power:
$$P = I^2 R = (1)^2 \times 100 = 100\ \text{W}$$

Cross-check via power factor:
- Total impedance magnitude: $$|Z| = \sqrt{R^2 + X_L^2} = \sqrt{100^2 + 100^2} = 141.4\ \Omega$$
- Total voltage: $$V = I|Z| = 1 \times 141.4 = 141.4\ \text{V}$$
- Apparent power: $$S = VI = 141.4\ \text{VA}$$
- Power factor: $$\cos\phi = \frac{R}{|Z|} = \frac{100}{141.4} = 0.707$$
- Real power: $$P = S\cos\phi = 141.4 \times 0.707 \approx 100\ \text{W}$$

Why the correct answer (100 W) is correct:
- The inductor (reactance) does not consume real power; only the resistor does. With 1 A through 100 Ω, $P = 100$ W.

Why the other choices are incorrect:
- 70.7 W: Results from misapplying power factor or using the wrong voltage; it is not the real power dissipated in the resistor.
- 141.4 W: This is the apparent power $S$, not the real power.
- 200 W: Treats reactance like resistance or adds series voltages arithmetically and uses $P=VI$ without considering phase.

Useful formulas and rules:
- Only resistors dissipate real power: $$P = I^2 R = \frac{V_R^2}{R}$$
- Apparent power: $$S = VI$$
- Impedance magnitude (series $R$ and $X$): $$|Z| = \sqrt{R^2 + X^2}$$
- Power factor in series $R$–$X$: $$\cos\phi = \frac{R}{|Z|}$$
- Reactive power (not dissipative): $$Q = I^2 X$$