Chapter 30, Problem 59P

(II) In the LRC circuit or Fig. 30–19, suppose I = I₀ sin ωt and V − V₀ sin(ωt + ϕ). Determine the instantaneous power dissipated in the circuit from P = IV using these equations and show that on the average, $\bar{P}=\frac{1}{2}{V}_0{I}_0\cos \varphi$, which confirms Eq. 30–30.

FIGURE 30-19 An LRC circuit.

We multiply the instantaneous current by the instantaneous voltage to calculate the instantaneous power. Then using the trigonometric identity for the summation of sine arguments (inside back cover of text) we can simplify the result. We integrate the power over a full period and divide the result by the period to calculate the average power.

P = IV = (I₀ sin ωt)V₀ sin(ωt + ϕ) = I₀V₀ sin ωt cosϕ +sinϕ cos ωt)

= I₀V₀ (sin² ωt cosϕ +sin ωt cos ωt sinϕ

$\begin{array}{l}\bar{P}\text{=}\frac{1}{T}{\int }_0^{T}PdT=\frac{\omega }{2\pi }{\int }_0^{\frac{2\pi }{\omega }}{I}_0{V}_0({\sin }^{\text{2}}+\omega t\cos \varphi +\sin \omega t\cos \omega t\sin \varphi )dt\\ =\frac{\omega }{2\pi }{I}_0{V}_0\cos \varphi {\int }_0^{\frac{2\pi }{\omega }}{\sin }^{\text{2}}\omega tdt+\frac{\omega }{2\pi }{I}_0{V}_0\sin \varphi {\int }_0^{\frac{2\pi }{\omega }}\sin \omega t\cos \omega tdt\\ =\frac{\omega }{2\pi }{I}_0{V}_0\cos \varphi \left(\frac{1}{2}\frac{2\pi }{\omega }\right)+\frac{\omega }{2\pi }{I}_0{V}_0\sin \varphi \left(\frac{1}{\omega }{\sin }^{\text{2}}\omega t{\int }_0^{\frac{2\pi }{\omega }}\right)=\frac{1}{2}{I}_0{V}_0\cos \varphi \end{array}$