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Answers without the blur. Sign up and see all textbooks for free! Q 3.5-6E

Expert-verified Found in: Page 121 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # 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.

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

See the step by step solution

## 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

${\mathbf{L}}\frac{\mathbf{dI}}{\mathbf{dt}}\mathbf{+}\mathbf{RI}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{LI}}\frac{\mathbf{dI}}{\mathbf{dt}}{\mathbf{+}}{{\mathbf{RI}}}^{{\mathbf{2}}}\mathbf{=}\mathbf{IE}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{d}}{\mathbf{dt}}{\mathbf{\left(}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{LI}}}^{{\mathbf{2}}}\mathbf{\right)}\mathbf{+}\mathbf{R}{{\mathbf{I}}}^{{\mathbf{2}}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{I}\phantom{\rule{0ex}{0ex}}$

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.

${\mathbf{R}}\frac{\mathbf{dq}}{\mathbf{dt}}{\mathbf{+}}\frac{\mathbf{q}}{\mathbf{c}}{\mathbf{=}}{\mathbf{E}}\phantom{\rule{0ex}{0ex}}{\mathbf{I}}{\mathbf{=}}\frac{\mathbf{dq}}{\mathbf{dt}}\phantom{\rule{0ex}{0ex}}{\mathbf{RI}}{\mathbf{+}}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{{\mathbf{RI}}}^{{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{I}\phantom{\rule{0ex}{0ex}}$

Further, solve the above expression

${\mathbf{q}}{\mathbf{=}}{{\mathbf{CE}}}_{{\mathbf{C}}}\phantom{\rule{0ex}{0ex}}{{\mathbf{RI}}}^{{\mathbf{2}}}{\mathbf{+}}\frac{{\mathbf{CE}}_{\mathbf{C}}}{\mathbf{C}}{\mathbf{.}}\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{I}\phantom{\rule{0ex}{0ex}}{{\mathbf{RI}}}^{{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{d}}{\mathbf{dt}}{\mathbf{\left(}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}{{\mathbf{E}}_{\mathbf{C}}}^{{\mathbf{2}}}\mathbf{\right)}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{I}\phantom{\rule{0ex}{0ex}}$

Therefore, the power balance equation for RC circuit is ${{\mathbf{RI}}}^{{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{d}}{\mathbf{dt}}\left(\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{{\mathbf{E}}_{\mathbf{C}}}^{\mathbf{2}}\right){\mathbf{=}}{\mathbf{E}}{ }\mathbf{\left(}\mathbf{t}\mathbf{\right)}{ }{\mathbf{I}}$ ### Want to see more solutions like these? 