recall and use the fact that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current

⚡️ Characteristics of Alternating Currents

In a sinusoidal alternating current (AC) circuit, the current and voltage vary as sine waves.

One key fact that you’ll need for the exam is that the mean (average) power delivered to a purely resistive load is exactly half the maximum power that can be delivered at the peak of the waveform.

📐 Key Formula

\$P{\text{mean}} = \frac{1}{2}\,P{\text{max}}\$

Where

• \$P{\text{max}} = I{\text{max}}^2 R\$ – the power when the current is at its peak value \$I_{\text{max}}\$.
• \$P{\text{mean}} = I{\text{rms}}^2 R\$ – the power averaged over one full cycle.

Since \$I{\text{rms}} = \dfrac{I{\text{max}}}{\sqrt{2}}\$ for a sine wave, the two expressions are equivalent.

🔌 Step‑by‑Step Derivation

1. Write the instantaneous current: \$I(t) = I_{\text{max}}\sin(\omega t)\$.
2. Instantaneous power into a resistor \$R\$: \$P(t) = I^2(t)R = I_{\text{max}}^2 R \sin^2(\omega t)\$.
3. Use the identity \$\sin^2x = \tfrac{1}{2}(1-\cos 2x)\$ to rewrite:

   \$P(t) = \tfrac{I_{\text{max}}^2 R}{2}\bigl(1-\cos(2\omega t)\bigr).\$

4. Average \$P(t)\$ over one full cycle (the cosine term averages to zero):

   \$P{\text{mean}} = \tfrac{I{\text{max}}^2 R}{2} = \tfrac{1}{2}P_{\text{max}}.\$

💡 Analogy: The Water‑Pipe Example

Imagine a water pipe that fills and empties rhythmically.

The maximum flow rate is the peak of the wave, but because the pipe is only full half the time, the *average* flow is half the maximum.

Similarly, AC power oscillates, so the average (mean) power is half the peak power.

📚 Example Problem

A 5 Ω resistor is connected to a 10 A peak sinusoidal current.

• Maximum power: \$P{\text{max}} = I{\text{max}}^2 R = 10^2 \times 5 = 500\ \text{W}\$.
• Mean power: \$P{\text{mean}} = \tfrac{1}{2}P{\text{max}} = 250\ \text{W}\$.

The same result can be found using RMS: \$I{\text{rms}} = 10/\sqrt{2} \approx 7.07\$ A, so \$P{\text{mean}} = I_{\text{rms}}^2 R \approx 7.07^2 \times 5 \approx 250\ \text{W}\$.

📝 Examination Tips

• Always check whether the problem asks for mean or maximum power.
• Remember the key relation: \$P{\text{mean}} = \dfrac{I{\text{rms}}^2 R}{1} = \dfrac{1}{2}P_{\text{max}}\$.
• When given a peak current, convert to RMS with \$I{\text{rms}} = I{\text{max}}/\sqrt{2}\$ before computing power.
• Show the steps of the derivation if the question asks for a short explanation.
• Use the water‑pipe analogy to explain why the average is half the peak – it’s a quick way to demonstrate understanding.

📊 Quick Reference Table