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Use differential equations to prove a formula for the current in a circuit

by RoRi

Consider an electric circuit as in Example 5 (page 318 of Apostol). Assume the voltage at time t is given by

    \[ V(t) = E \sin \omega t, \]

where E and \omega are positive constants. If Z(0) = 0, prove that the current at time t is given by

    \[ I(t) = \frac{E}{\sqrt{R^2 + \omega^2 L^2}} \sin (\omega t - \alpha) + \frac{E \omega L}{R^2 + \omega^2 L^2} e^{-Rt/L}, \]

where \alpha depends only on \omega, L, and R. Prove that a = 0 when L = 0.

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

    \[ LI'(t) + RI(t) = V(t) \quad \implies \quad I(t) = I(0) e^{-\frac{Rt}{L}} + e^{-\frac{Rt}{L}} \int_0^t \frac{V(x)}{L} e^{\frac{Rx}{L}} \, dx. \]

So, if V(t) = E \sin (\omega t) and I(0) = 0, where E and \omega are positive constants, we have

    \begin{align*} I(t) &= e^{-\frac{Rt}{L}} \int_0^t \frac{E \sin (\omega t)}{L} e^{\frac{Rx}{L}} \, dx \\[9pt] &= \frac{e^{-\frac{Rt}{L}} \cdot E}{L} \left( \frac{RL \sin (\omega t) \cdot e^{\frac{Rt}{L}} - L^2 \omega \cos (\omega t) \cdot e^{\frac{Rt}{L}}}{R^2 + \omega^2 L^2} + \frac{L^2 \omega}{R^2 + \omega^2 L^2} \right) \\[9pt] &= \frac{ER\sin (\omega t) - E \omega L \cos (\omega t)}{R^2 + \omega^2 L^2} + \frac{E \omega L}{R^2 + \omega^2 L^2} \cdot e^{-\frac{Rt}{L}}. \end{align*}

Then, let

    \[ \sin \alpha = \frac{\omega L}{\sqrt{R^2+\omega^2 L^2}}, \qquad \cos \alpha = \frac{R}{\sqrt{R^2+\omega^2 L^2}}. \]

(By the Pythagorean identity we can make this choice.) Therefore,

    \begin{align*} I(t) &= \left( \frac{E}{\sqrt{R^2+\omega^2 L^2}} \right) \left( \cos \alpha \sin (\omega t) - \sin \alpha \cos (\omega t) \right) + \frac{E \omega L}{R^2 + \omega^2 L^2} \cdot e^{-\frac{Rt}{L}} \\[9pt] &= \left( \frac{E}{\sqrt{R^2 + \omega^2 L^2}} \right) \left( \sin (\omega t - \alpha) \right) + \frac{E \omega L}{R^2 + \omega^2 L^2} \cdot e^{-\frac{Rt}{L}}. \qquad \blacksquare \end{align*}

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