An electrical component has voltage drop v = V_msin(ωt), when the current through it is i = I_msin(ωt − θ). What is the average power dissipated over a half cycle corresponding to ω?

The correct option is: (V_mI_m/2)cosθ

Step-by-Step Derivation and Explanation:

1. Understand the Expressions:
We are given the voltage drop $v$ and the current $i$ as functions of time $t$:
$v\left(t\right)=V_{m}sin\left(\omega t\right)$
$i\left(t\right)=I_{m}sin(\omega t-\theta )$

2. Express the Instantaneous Power:
Instantaneous power $p\left(t\right)$ is the product of the instantaneous voltage and instantaneous current:

$$p\left(t\right)=v\left(t\right)·i\left(t\right)=V_m{I}_{m}sin\left(\omega t\right)sin(\omega t-\theta )$$

3. Apply Trigonometric Identity:
Using the product-to-sum identity:

$$sinAsinB=\frac{1}{2}[cos(A-B)-cos(A+B\left)\right]$$

Let $A=\omega t$ and $B=\omega t-\theta$. This yields:

$$p\left(t\right)=\frac{V_m{I}_m}{2}[cos\theta -cos(2\omega t-\theta \left)\right]$$

4. Calculate the Average Power over a Half Cycle:
For a frequency of $\omega$, a full period is $T=\frac{2\pi }{\omega }$. Therefore, the duration of a half cycle is:

$$t=\frac{T}{2}=\frac{\pi }{\omega }$$

The average value of the power $P_{\text{avg}}$ over this half-cycle interval from $t=0$ to $t=\frac{\pi }{\omega }$ is:

$$P_{\text{avg}}=\frac{1}{\pi /\omega }$$

Substituting $p\left(t\right)$:

$$P_{\text{avg}}=\frac{\omega }{\pi }{\int }_0^{\pi /\omega }\frac{V_m{I}_m}{2}[cos\theta -cos(2\omega t-\theta \left)\right]dt$$

5. Evaluate the Integrals:
The term $cos(2\omega t-\theta )$ is a periodic sinusoidal function with angular frequency $2\omega$. Its complete period is:

$$T_p=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$$

Integrating any sinusoidal wave over one full period of its own oscillation evaluates to zero:

$${\int }_0^{\pi /\omega }cos(2\omega t-\theta )dt=0$$

Thus, only the constant term $cos\theta$ contributes to the average power:

$$P_{\text{avg}}=\frac{\omega }{\pi }·\frac{V_m{I}_m}{2}cos\theta {\int }_0^{\pi /\omega }dt$$

$$P_{\text{avg}}=\frac{\omega }{\pi }·\frac{V_m{I}_m}{2}cos\theta ·\left(\frac{\pi }{\omega }\right)$$

Simplifying by canceling $\frac{\omega }{\pi }$ and $\frac{\pi }{\omega }$:

$$P_{\text{avg}}=\frac{V_m{I}_m}{2}cos\theta$$