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

Learning Objective

Recall and use the fact that the mean (average) power delivered to a purely resistive load by a sinusoidal alternating current is half the maximum (peak) power.

Key Concepts

• Sinusoidal alternating current (AC) can be described by \$i(t)=I{\max}\sin(\omega t)\$, where \$I{\max}\$ is the peak current.
• The corresponding voltage across a resistor \$R\$ is \$v(t)=V{\max}\sin(\omega t)\$ with \$V{\max}=I_{\max}R\$.
• Instantaneous power in a resistor is \$p(t)=v(t)i(t)=\frac{v^{2}(t)}{R}= \frac{i^{2}(t)R}{ }\$.
• Mean (average) power over one full cycle is denoted \$\overline{P}\$.

Derivation of Mean Power for a Sinusoidal AC

Starting from the instantaneous power expression for a resistor:

\$p(t)=v(t)i(t)=V{\max}I{\max}\sin^{2}(\omega t)\$

Using the trigonometric identity \$\sin^{2}x=\frac{1-\cos 2x}{2}\$, we obtain:

\$p(t)=\frac{V{\max}I{\max}}{2}\bigl[1-\cos(2\omega t)\bigr]\$

The mean power \$\overline{P}\$ is the average of \$p(t)\$ over one period \$T=\frac{2\pi}{\omega}\$:

\$\$\overline{P}= \frac{1}{T}\int_{0}^{T}p(t)\,dt

=\frac{V{\max}I{\max}}{2}\left[\frac{1}{T}\int{0}^{T}1\,dt-\frac{1}{T}\int{0}^{T}\cos(2\omega t)\,dt\right]\$\$

The integral of \$\cos(2\omega t)\$ over a full cycle is zero, leaving:

\$\boxed{\;\overline{P}= \frac{V{\max}I{\max}}{2}\;}\$

Since the maximum (peak) power occurs when \$\sin^{2}(\omega t)=1\$, the peak power is

\$P{\max}=V{\max}I_{\max}\$

Therefore, for a sinusoidal AC in a resistive load:

\$\boxed{\;\overline{P}= \frac{1}{2}P_{\max}\;}\$

RMS (Root‑Mean‑Square) Values

RMS values provide a convenient way to express AC quantities in terms of equivalent DC values:

\$I{\text{rms}}=\frac{I{\max}}{\sqrt{2}},\qquad V{\text{rms}}=\frac{V{\max}}{\sqrt{2}}\$

Using RMS values, the mean power can also be written as:

\$\overline{P}=I{\text{rms}}^{2}R = \frac{V{\text{rms}}^{2}}{R}=V{\text{rms}}I{\text{rms}}\$

These expressions are algebraically equivalent to the “half‑maximum” result derived above.

Summary Table

Worked Example

Problem: A sinusoidal AC source supplies a resistor \$R=10\;\Omega\$. The peak voltage across the resistor is \$V_{\max}=100\;\text{V}\$. Find the mean power dissipated in the resistor.

1. Calculate the RMS voltage:

   \$V{\text{rms}}=\frac{V{\max}}{\sqrt{2}}=\frac{100}{\sqrt{2}}\;\text{V}\approx 70.7\;\text{V}\$

2. Use the RMS formula for power:

   \$\overline{P}= \frac{V_{\text{rms}}^{2}}{R}= \frac{(70.7)^{2}}{10}\;\text{W}\approx 500\;\text{W}\$

3. Check using the “half‑maximum” rule:

   \$P{\max}= \frac{V{\max}^{2}}{R}= \frac{100^{2}}{10}=1000\;\text{W}\$

   \$\overline{P}= \frac{1}{2}P_{\max}= \frac{1}{2}\times1000=500\;\text{W}\$

The two methods give the same result, confirming the relationship.

Common Misconceptions

• “Average” vs “RMS”: The arithmetic average of a sinusoidal voltage over a full cycle is zero, not the useful value for power calculations. RMS is the appropriate average for power.
• Peak vs RMS values: Confusing \$V{\max}\$ with \$V{\text{rms}}\$ leads to errors of a factor of \$\sqrt{2}\$ in power calculations.
• Non‑resistive loads: The half‑maximum rule only holds for purely resistive loads. In inductive or capacitive circuits, phase differences modify the average power.

Practice Questions

1. For a sinusoidal current with \$I_{\max}=8\;\text{A}\$ flowing through a \$5\;\Omega\$ resistor, calculate:

   • Peak power \$P_{\max}\$
   • Mean power \$\overline{P}\$
   • RMS current \$I{\text{rms}}\$ and RMS power \$P{\text{rms}}\$

2. A household appliance is rated at \$1200\;\text{W}\$ mean power when connected to a \$240\;\text{V}\$ RMS supply. Assuming the appliance behaves as a pure resistor, determine its resistance and the peak voltage across it.
3. Explain why the mean power delivered to a purely capacitive load is zero, even though the instantaneous power is non‑zero during part of each cycle.