Q 3.5-5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The power generated or dissipated by a circuit element equals the voltage across the element times the current through the element. Show that the power dissipated by a resistor equal  l²R, the power associated with an inductor equals the derivative of $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{LI}}^{\mathbf{2}}$  and the power associated with a capacitor equals the derivative of  $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{{\mathbf{E}}_{\mathbf{c}}}^{\mathbf{2}}$.

Step-by-Step Solution

Answer

• The power dissipated by a circuit is  ${\mathbf{P}}_{\mathbf{R}}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{R}$ .

 

• The power dissipated by an inductor is  ${\mathbf{P}}_{\mathbf{L}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{LI}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]$

 

• The power associated by a capacitor is   ${\mathbf{P}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{dt}}\left[\mathbf{C}{{\mathbf{E}}_{\mathbf{C}}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]$

Step 1: Evaluate the power dissipated by a circuit

The power is given by the equation P = I(t)V(t).

 

Since the voltage across resister is 

 $$\begin{gathered} \mathrm{V}\left(\mathrm{t}\right)={\mathrm{E}}_{\mathrm{R}}\left(\mathrm{t}\right)=\mathrm{RI}\left(\mathrm{t}\right) \\ {\mathrm{P}}_{\mathrm{R}}=\mathrm{I}\left(\mathrm{t}\right)\mathrm{RI}\left(\mathrm{t}\right) \\ {\mathrm{P}}_{\mathrm{R}}={\mathrm{I}}^2\left(\mathrm{t}\right)\mathrm{R} \end{gathered}$$

 

Hence, the power dissipated by a circuit is ${\mathbf{P}}_{\mathbf{R}}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{R}$ .

Step 2: Find the power dissipated by an inductor.

Since the voltage across the inductor is 

$$\begin{gathered} \mathrm{V}\left(\mathrm{t}\right)={\mathrm{E}}_{\mathrm{L}}\left(\mathrm{t}\right) \\ =\mathrm{L}\frac{\mathrm{dI}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ {\mathrm{P}}_{\mathrm{L}}=\mathrm{I}\left(\mathrm{t}\right)\mathrm{L}\frac{\mathrm{dI}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ =\frac{\mathrm{L}}{2}2\mathrm{I}\left(\mathrm{t}\right)\frac{\mathrm{dI}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ {\mathrm{P}}_{\mathrm{L}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{{\mathrm{LI}}^2\left(\mathrm{t}\right)}{2}\right] \\ {\mathrm{P}}_{\mathrm{L}}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{LI}}^2\left(\mathrm{t}\right)\right] \end{gathered}$$

Hence, the power dissipated by an inductor is  ${\mathbf{P}}_{\mathbf{L}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{LI}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]$

Step 3: Determine the power associated by a capacitor

Since the voltage across the capacitor is 

$$\begin{gathered} \mathrm{V}\left(\mathrm{t}\right)={\mathrm{E}}_{\mathrm{C}}\left(\mathrm{t}\right) \\ =\frac{1}{\mathrm{C}}\mathrm{q}\left(\mathrm{t}\right) \\ \mathrm{q}\left(\mathrm{t}\right)=\mathrm{C}{\mathrm{E}}_{\mathrm{c}}\left(\mathrm{t}\right) \\ \mathrm{I}\left(\mathrm{t}\right)=\frac{\mathrm{dq}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ =\frac{\mathrm{d}\left[{\mathrm{CE}}_{\mathrm{c}}\left(\mathrm{t}\right)\right]}{\mathrm{dt}} \\ {\mathrm{P}}_{\mathrm{C}}=\frac{\mathrm{d}\left[{\mathrm{CE}}_{\mathrm{C}}\left(\mathrm{t}\right)\right]}{\mathrm{dt}}\frac{\mathrm{q}\left(\mathrm{t}\right)}{\mathrm{C}} \\ =\frac{\mathrm{C}}{2}2{\mathrm{E}}_{\mathrm{C}}\left(\mathrm{t}\right)\frac{{\mathrm{dE}}_{\mathrm{C}}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ {\mathrm{P}}_{\mathrm{C}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\mathrm{C}{{\mathrm{E}}_{\mathrm{C}}}^2\left(\mathrm{t}\right)}{2}\right] \\ {\mathrm{P}}_{\mathrm{C}}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{C}{{\mathrm{E}}_{\mathrm{C}}}^2\left(\mathrm{t}\right)\right] \end{gathered}$$

Hence, the power associated by a capacitor is  ${\mathbf{P}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{d}}{\mathbf{dt}}\left[\mathbf{C}{{\mathbf{E}}_{\mathbf{C}}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]$