Q 3.5-4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

If the resistance in the RL circuit of Figure 3.13(a) is zero, show that the current I (t) is directly proportional to the integral of the applied voltage E(t). Similarly, show that if the resistance in the RC circuit of Figure 3.13(b) is zero, the current is directly proportional to the derivative of the applied voltage.

Step-by-Step Solution

Answer

 Proved

Step 1: Show that I is directly proportional to the integral of the applied voltage

Apply Kirchhoff’s voltage law for RL circuit is  $\frac{\mathbf{dI}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{RI}}{\mathbf{L}}\mathbf{=}\frac{\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{L}}$.

 

When R = 0 then equation becomes.

 $$\begin{gathered} \frac{\mathbf{dI}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{L}} \\ \int \mathbf{dI}\mathbf{=}\int \frac{\mathbf{1}}{\mathbf{L}}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt} \\ \mathbf{I}\mathbf{=}\int \frac{\mathbf{1}}{\mathbf{L}}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt} \end{gathered}$$

 

Hence it is proved that the current I (t) is directly proportional to the integral of the applied voltage E(t).

Step 2: Evaluate current is directly proportion to the derivative of the applied voltage.

Apply Kirchhoff’s voltage law for RL circuit is   $\mathbf{RI}\mathbf{+}\frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

When R = 0 then equation is 

 $$\begin{gathered} \frac{\mathbf{q}}{\mathbf{C}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{q}\mathbf{=}\mathbf{CE}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{I}\mathbf{=}\frac{\mathbf{dq}}{\mathbf{dt}} \\ \mathbf{I}\mathbf{=}\frac{\mathbf{d}\mathbf{\left(}\mathbf{CE}\mathbf{\right(}\mathbf{t}\mathbf{)}}{\mathbf{dt}} \\ \mathbf{I}\mathbf{=}\mathbf{C}\frac{\mathbf{d}\mathbf{\left(}\mathbf{E}\mathbf{\right(}\mathbf{t}\mathbf{)}}{\mathbf{dt}} \end{gathered}$$

 

Hence it is proved that the current is directly proportional to the derivative of the applied voltage.