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: Write the instantaneous power.
The instantaneous power absorbed by the component is
\[ p(t) = v(t)\,i(t) = V_mI_m\sin(\omega t)\sin(\omega t-\theta) \]

Step 2: Convert the product of sines to a sum.
Using \(\sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]\) with \(A=\omega t\) and \(B=\omega t-\theta\),
\[ p(t) = \frac{V_mI_m}{2}\Big[\cos\theta-\cos(2\omega t-\theta)\Big] \]

Step 3: Notice the two parts of this expression.
The first part, \(\frac{V_mI_m}{2}\cos\theta\), does not depend on time at all, it is a constant.
The second part, \(-\frac{V_mI_m}{2}\cos(2\omega t-\theta)\), oscillates at frequency \(2\omega\), which is twice the frequency of \(v\) and \(i\). Its own period is
\[ \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]
which is exactly half the period of \(v\) and \(i\) (their period is \(2\pi/\omega\)).

Step 4: Average \(p(t)\) over a half cycle.
A half cycle of \(v\) and \(i\) lasts \(\pi/\omega\) seconds, which is precisely one full period of the oscillating term \(\cos(2\omega t-\theta)\). The average of a cosine over exactly one of its own full periods is always zero, no matter where that interval starts. So over any half cycle,
\[ \text{average of }\cos(2\omega t-\theta) = 0 \]

Step 5: Find the average power.
Only the constant part survives the averaging:
\[ P_{avg} = \frac{V_mI_m}{2}\cos\theta-0 = \frac{V_mI_m}{2}\cos\theta \]

Step 6: Compare with the full-cycle average.
This is worth noting: the average power over a half cycle comes out to be exactly the same as the average power over one full cycle, since the ripple term always completes a whole number of its own periods within any half cycle of \(v\) and \(i\). This rules out choice (A) (which would only hold if we averaged over a special quarter-cycle window, not a genuine half cycle), and choices (B) and (D), which correspond to using peak values directly or dividing by an extra factor of \(2\) by mistake.

Final Answer:
\[ \boxed{P_{avg} = \frac{V_mI_m}{2}\cos\theta} \]