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Use differential equations to determine the current in a circuit

by RoRi

This exercise refers to the electric circuit in Example 5 on page 318 of Apostol. Assume the voltage in the circuit at time t is given by

    \[ E(t) = E \qquad \text{if} \qquad a \leq t \leq b, \]

and

    \[ E(t) = 0 \qquad \text{for all other } t. \]

If I(0) = 0 prove the current at time t is given by

    \[ I(t)= 0 \qquad \text{if} \qquad t \leq a; \]

and

    \[ I(t) = \frac{E}{R}\left( 1- e^{-R(t-a)/L} \right) \qquad \text{if} \qquad a \leq t \leq b; \]

and

    \[ I(t) = \frac{E}{R} e^{-\frac{Rt}{L}} \left( e^{-\frac{Rb}{L}} - e^{-\frac{Ra}{L}}\right) \qquad \text{if} \qquad t \geq b. \]

Proof. By Kirchhoff’s law (page 317 of Apostol) we have

    \[ I(t) = I(0) e^{-\frac{Rt}{L}} + e^{-\frac{Rt}{L}} \int_0^t \frac{V(x)}{L} e^{\frac{Rx}{L}} \, dx. \]

So, for I(0) = 0 and t \leq a we have V(t) = E(t) = 0 and I(t) = 0. If a \leq t \leq b then I(a) = 0 and E(t) = E so we have

    \begin{align*} I(t) &= e^{-\frac{Rt}{L}} \int_a^t \frac{E}{L} e^{\frac{Rx}{L}} \, dx \\[9pt] &= e^{-\frac{Rt}{L}} \cdot \frac{E}{R} \left( e^{\frac{Rt}{L}} - e^{\frac{Ra}{L}} \right) \\[9pt] &= \frac{E}{R} \left( 1 - e^{-\frac{R(t-a)}{L}} \right). \end{align*}

Finally, if t \geq b then I(b) = \frac{E}{R} \left(1 - e^{-\frac{R(b-a)}{L}} \right) and E(t) = 0, so

    \begin{align*} I(t) &= \frac{E}{R} \left( 1 - e^{-\frac{R(b-a)}{L}} \right) e^{-\frac{Rt}{L}} \\[9pt] &= \frac{E}{R} e^{-\frac{Rt}{L}} \left( e^{\frac{Rb}{L}} - e^{\frac{Ra}{L}} \right). \qquad \blacksquare \end{align*}

A sketch of I(t) where we let a = 1, b = 2, and E = R = 1:

Rendered by QuickLaTeX.com

5 comments

  2. tom says:

    Just want to point out the graph looks suspicious. Doesn’t an inductor resist change in current, then attempts to maintain current when the source voltage is removed? So how can there be a negative current flow? Also, the term E/R(1-e^-(R(t-a)/L) can never be negative, being E is positive and e^-(R(t-a)/L is < 1.

    • tom says:

      Never mind :( If I looked carefully I would have seen the equations match the graph, but reading about the counter electromotive force set me straight.

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