Step 1: Understanding the Problem:
The question asks for the average (real) power consumed by a load impedance when an AC voltage is applied to it.
The load impedance consists of both a resistive part and a reactive part, but only the resistive component dissipates real average power.
Step 2: Key Formula or Approach:
1. Convert the time-domain voltage to its RMS value:
\[ V_{rms} = \frac{V_m}{\sqrt{2}} \]
2. Calculate the magnitude of the load impedance (\(Z = R + jX\)):
\[ |Z| = \sqrt{R^2 + X^2} \]
3. Compute the RMS current (\(I_{rms}\)) flowing through the impedance:
\[ I_{rms} = \frac{V_{rms}}{|Z|} \]
4. Calculate the average power (\(P_{avg}\)) dissipated in the resistor:
\[ P_{avg} = I_{rms}^2 \cdot R \]
Step 3: Detailed Explanation:
• Given the voltage equation \(v(t) = 50\sqrt{2} \sin(\omega t)\), the peak voltage is \(V_m = 50\sqrt{2}\text{ V}\).
• Calculate the RMS voltage:
\[ V_{rms} = \frac{50\sqrt{2}}{\sqrt{2}} = 50\text{ V} \]
• Given the impedance \(Z = 3 + j4\ \Omega\), identify the resistive and reactive parts:
- Resistance, \(R = 3\ \Omega\).
- Inductive Reactance, \(X_L = 4\ \Omega\).
• Calculate the magnitude of the impedance:
\[ |Z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\ \Omega \]
• Calculate the RMS current:
\[ I_{rms} = \frac{V_{rms}}{|Z|} = \frac{50}{5} = 10\text{ A} \]
• Calculate the average power dissipated in the load:
\[ P_{avg} = I_{rms}^2 \cdot R = (10)^2 \times 3 = 100 \times 3 = 300\text{ W} \]
Step 4: Final Answer:
The average power is 300 W.