Q 3.5-6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Derive a power balance equation for the RL and RC circuits. (See Problem 5.) Discuss the significance of the signs of the three power terms.

Step-by-Step Solution

Answer

The power balance equation for RL and RC circuit are  $\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{LI}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\mathbf{R}{\mathbf{I}}^{\mathbf{2}}\mathbf{=}\mathbf{E} \mathbf{\left(}\mathbf{t}\mathbf{\right)} \mathbf{I}$ and  ${\mathbf{RI}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{d}}{\mathbf{dt}} \left(\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{{\mathbf{E}}_{\mathbf{C}}}^{\mathbf{2}}\right)\mathbf{=}\mathbf{E} \mathbf{\left(}\mathbf{t}\mathbf{\right)} \mathbf{I}$ respectively.

Step 1: Important formula.

The governing differential equation for RC circuit is $\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{q}}{\mathbf{CR}}\mathbf{=}\frac{\mathbf{E}}{\mathbf{R}}$ .

Step 2: Determine the power balance for RL circuit

The power dissipated by a circuit is  ${\mathbf{P}}_{\mathbf{R}}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \mathbf{R}$ and the power dissipated by an inductor is ${\mathbf{P}}_{\mathbf{L}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{ }\mathbf{ }\left[{\mathrm{LI}}^2\left(t\right)\right]$  and the power associated by a capacitor is ${\mathbf{P}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{dt}}\left[C{E_C}^2\left(t\right)\right]$ .

 

So, the equation is

$$\begin{gathered} \mathbf{L}\frac{\mathbf{dI}}{\mathbf{dt}}\mathbf{+}\mathbf{RI}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{LI}\frac{\mathbf{dI}}{\mathbf{dt}}\mathbf{+}{\mathbf{RI}}^{\mathbf{2}}\mathbf{=}\mathbf{IE}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dt}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{LI}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\mathbf{R}{\mathbf{I}}^{\mathbf{2}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{I} \end{gathered}$$

Therefore, the power balance equation for RL circuit is $\frac{\mathbf{d}}{\mathbf{dt}}\left(\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{LI}}^{\mathbf{2}}\right)\mathbf{+}{\mathbf{RI}}^{\mathbf{2}}\mathbf{=}\mathbf{E} \mathbf{\left(}\mathbf{t}\mathbf{\right)} \mathbf{I}$ .

Step 3: evaluate the power balance for RC circuit.

 $$\begin{gathered} \mathbf{R}\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{q}}{\mathbf{c}}\mathbf{=}\mathbf{E} \\ \mathbf{I}\mathbf{=}\frac{\mathbf{dq}}{\mathbf{dt}} \\ \mathbf{RI}\mathbf{+}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ {\mathbf{RI}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{I} \end{gathered}$$

 Further, solve the above expression

 

 $$\begin{gathered} \mathbf{q}\mathbf{=}{\mathbf{CE}}_{\mathbf{C}} \\ {\mathbf{RI}}^{\mathbf{2}}\mathbf{+}\frac{{\mathbf{CE}}_{\mathbf{C}}}{\mathbf{C}}\mathbf{.}\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{I} \\ {\mathbf{RI}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{{\mathbf{E}}_{\mathbf{C}}}^{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{E}\mathbf{(}\mathbf{t}\mathbf{)}\mathbf{I} \end{gathered}$$

 

Therefore, the power balance equation for RC circuit is  ${\mathbf{RI}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{d}}{\mathbf{dt}}\left(\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{{\mathbf{E}}_{\mathbf{C}}}^{\mathbf{2}}\right)\mathbf{=}\mathbf{E} \mathbf{\left(}\mathbf{t}\mathbf{\right)} \mathbf{I}$