Question

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.

Short answer

a) The energy stored in the capacitor does not change.

b) The energy stored in the inductor does not change.

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

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

e) The energy supplied by driving emf is equal to the energy dissipated by the resistor.

Step-by-step solution

Step 1: Given

An RLC circuit with period$T.$

Determining the concept

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:

• Energy stored in the capacitor, $U_E=\frac{q^2}{\text{2}C}$.
• Energy stored in the inductor,.$U_B=\frac{\text{1}}{\text{2}}L{i}^2$
• Rate of energy supply by driving emf device, $P_{ϵ}=iϵ$.Where, $i=I\text{sin}({\omega }_d-\varphi )$and .$ϵ={ϵ}_m\text{sin}{\omega }_{d}t$
• Rate with which energy dissipates, $P_R=i^{2}R$

(a) Determining the energy stored in the capacitor does not change

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.

(b) Determining the energy stored in the inductor 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.

(c) Determining the driving emf device supplies energy  (12T)ϵmI cos ϕ

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.$\left(\frac{\text{1}}{\text{2}}T\right){ϵ}_{m}Icos\varphi$

(d) Determining the resistor dissipates energy  (12T)RI2

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.$\left(\frac{\text{1}}{\text{2}}T\right)R{I}^2$

(e) Determining the quantities found in part c and part d are equal

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.