For a sinusoidally driven series RLCcircuit, show that over one complete cycle with period T (a) the energy stored in the capacitor does not change; (b) the energy stored in the inductor does not change; (c) the driving emf device supplies energy $\mathbf{\left(}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\mathbf{T}\mathbf{\right)}{\mathit{ϵ}}_{\mathbf{m}}\mathbf{I}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }$ and (d) the resistor dissipates energy.$\left(\frac{1}{2}\mathbf{T}\right){\mathbf{RI}}^2$(e) Show that the quantities found in (c) and (d) are equal.

An RLC circuit with period$T.$

By using the concept of energy stored in the capacitor, energy stored in the inductor, the energy supplied by the emf device, and the energy dissipated in the resistor, prove all the parts.

The formulae are as follows:

It is known, that charge$q$is a periodic function of$t$with period .$T$

As.$U_E=\frac{q^2}{\text{2}C}$

So $U_E$must also be the periodic function t.

Thus, the energy stored in the capacitor does not change.

Since the current$i$is also periodic$t$with the period .$T$

As,

$U_B=\frac{\text{1}}{\text{2}}L{i}^2$.

So,$U_B$must also be the periodic function .$t$

Thus, the energy stored in the inductor does not change.

The energy supplied by the emf device over one cycle is,

$\begin{array}{l}{U}_E=\underset{0}{\overset{T}{}}{P}_{ϵ}dt=I{ϵ}_m\underset{0}{\overset{T}{}}\text{sin}({\omega }_{d}t-\varphi )\text{sin}\left({\omega }_{d}t\right)dt\\ U_E=I{ϵ}_m\underset{0}{\overset{T}{}}\text{[sin}\left({\omega }_{d}t\right)\text{cos\hspace{0.17em}}\varphi -\text{cos}\left({\omega }_{d}t\right)\text{sin\hspace{0.17em}}\varphi ]\text{sin}\left({\omega }_{d}t\right)dt\end{array}$

Since,

And,

$\begin{array}{c}\underset{0}{\overset{T}{}}{\text{sin}}^2\left({\omega }_{d}t\right)dt=\frac{T}{\text{2}}\\ \underset{0}{\overset{T}{}}\text{sin}\left({\omega }_{d}t\right)\text{cos}\left({\omega }_{d}t\right)dt=\frac{T}{\text{2}}\end{array}$

Therefore,

$U_E=\frac{T}{\text{2}}I{ϵ}_m\text{cos\hspace{0.17em}}\varphi$.

Hence,energy supplied by driving emf is.$\left(\frac{\text{1}}{\text{2}}T\right){ϵ}_{m}Icos\varphi$

Over one cycle, the energy dissipated in the resistor is,

$U_R=\underset{0}{\overset{T}{}}{P}_{R}dt=I^{2}R\underset{0}{\overset{T}{}}{\text{sin}}^2({\omega }_{d}t-\varphi )dt$.

$U_R=\frac{T}{\text{2}}{I}^{2}R$.

Hence,energy dissipated by the resistor is.$\left(\frac{\text{1}}{\text{2}}T\right)R{I}^2$

From part c, the energy supplied by the emf device over one cycle is,

$\begin{array}{c}{U}_E=\frac{T}{\text{2}}I{ϵ}_m\text{cos\hspace{0.17em}}\varphi \\ U_E=\frac{T}{\text{2}}I{ϵ}_m\left(\frac{V_R}{{ϵ}_m}\right)\\ U_E=\frac{T}{\text{2}}I{ϵ}_m\left(\frac{IR}{{ϵ}_m}\right)\\ U_E=\frac{T}{\text{2}}{I}^{2}R\end{array}$.

This is equal to theenergy dissipated in the resistor from part d.

Hence, the quantities found in parts c and d are equal.

Hence,energy supplied by driving emf is equal to the energy dissipated by the resistor.

By using the concept of energy stored in the capacitor, energy stored in the inductor, the energy supplied by the emf device, and the energy dissipated in the resistor, all the results.