Question

An electrical component has voltage drop \(v = V_m\sin(\omega t)\), when the current through it is \(i = I_m\sin(\omega t-\theta)\).
What is the average power dissipated over a half cycle corresponding to \(\omega\)?

• A. \(0\)
• B. \(V_mI_m\cos\theta\)
• C. \(\dfrac{V_mI_m}{2}\cos\theta\)
• D. \(\dfrac{V_mI_m}{4}\cos\theta\)

Hint:

Split the product of sines into a constant term and a double-frequency ripple term; the ripple always averages to zero over a half cycle.

Answer 1

The Correct Option is C

Step 1: Set the half-cycle window.
Take the half cycle from $t=0$ to $t=\pi/\omega$ (any other half cycle gives the same result by symmetry). The average power over this window is
\[ P_{avg}=\frac{\omega}{\pi}\int_0^{\pi/\omega} V_mI_m\sin(\omega t)\sin(\omega t-\theta)\,dt \]

Step 2: Expand the current term.
Write $\sin(\omega t-\theta)=\sin(\omega t)\cos\theta-\cos(\omega t)\sin\theta$. Then
\[ p(t)=V_mI_m\big[\sin^2(\omega t)\cos\theta-\sin(\omega t)\cos(\omega t)\sin\theta\big] \]

Step 3: Average each piece over the half cycle.
Over a half cycle, $\sin^2(\omega t)$ averages to $\dfrac{1}{2}$ (the same as its average over a full cycle, since $\sin^2$ repeats twice as fast as $\sin$ itself). The term $\sin(\omega t)\cos(\omega t)=\dfrac12\sin(2\omega t)$ is a pure sine wave of frequency $2\omega$, and it averages to exactly $0$ over the half cycle $[0,\pi/\omega]$, which is one whole period of $\sin(2\omega t)$.

Step 4: Combine.
\[ P_{avg}=V_mI_m\left[\cos\theta\cdot\frac12-\sin\theta\cdot 0\right]=\frac{V_mI_m}{2}\cos\theta \]

Step 5: Compare with the other choices.
The plain product $V_mI_m\cos\theta$ (option B) forgets that both $v$ and $i$ are amplitude values, not rms values, so it is too large by a factor of $2$. Dividing by $4$ (option D) over-corrects. Zero (option A) would only apply to a special quarter-cycle window, not a genuine half cycle of $v$ and $i$.
\[ \boxed{P_{avg}=\dfrac{V_mI_m}{2}\cos\theta} \]