*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?*  

*Correct Answer is  B: 100 watts*  

### AI Response: (may contain errors)  

To determine the real power consumed in a circuit with a resistor and an inductive reactance, we need to focus on the power dissipated by the resistor, as inductors do not consume real power.

### Explanation:

1. **Real Power Calculation:**
   - Real power ($P$) is only consumed by the resistive part of the circuit.
   - The formula for real power is:  
     $$P = I^2 \times R$$  
     where:
     - $I = 1 \, \text{A}$ (current)
     - $R = 100 \, \Omega$ (resistance)

   - Substituting the values:  
     $$P = (1 \, \text{A})^2 \times 100 \, \Omega = 100 \, \text{W}$$

2. **Why the Correct Answer is 100 Watts:**
   - The real power is calculated using only the resistive component, which is 100 ohms. The inductive reactance does not contribute to real power consumption.

3. **Why Other Choices are Incorrect:**
   - **70.7 watts (Choice A):** This might be a confusion with apparent power or power factor calculations, but it does not apply here since we are only interested in real power.
   - **141.4 watts (Choice C):** This could be a result of misapplying the Pythagorean theorem for impedance, but it does not reflect real power.
   - **200 watts (Choice D):** This would be the apparent power if you incorrectly added the resistive and reactive components directly, but it is not the real power consumed.

### Key Points:
- Real power is only consumed by resistive elements.
- Inductive reactance contributes to reactive power, not real power.
- Use the formula $P = I^2 \times R$ to find real power in resistive components.