Question

# An $RL$ circuit with a $5-\Omega$ resistor and a $0.05$-H inductor carries a current of $1$ A at $t = 0,$ at which time a voltage source $E ( t) = 5 \cos 120t V$ is added. Determine the subsequent inductor current and voltage.

Solution

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$\textbf{Given:}$

$R=5 \; \Omega, \quad L=0.05 \; H, \quad I(t)=I(0)=1, \quad \text{and} \quad E(t)= 5\cos(120t) \;V$

First we determine the $\text{\underline{subsequent inductor current}}$, where,

\begin{align} I(t) &= e^{-Rt/L}\left[ \int e^{Rt/L} \dfrac{E(t)}{L}\;dt + K\right] \\ \\ I(t) &= e^{-5t/0.05} \left[ \int e^{5t/0.05} \dfrac{5\cos(120t)}{0.05} \;dt + K \right] \\ \\ I(t)&=e^{-100t} \left[ 100 \int e^{100t} \cos(120t) \;dt + K \right] \end{align}

To evaluate the integral in $(1)$, we will use integration by parts, where,

\begin{align*} \int e^{100t} \cos(120t) \;dt \end{align*}

\begin{align*} u&=\cos(120t) & dv&=e^{100t} \\ du&=-120 \sin(120t) & v&=\dfrac{e^{100t}}{100} \end{align*}

and so,

\begin{align} \int e^{100t} \cos(120t) \;dt &= \dfrac{e^{100t}\cos(120t)}{100}+\int 120 \sin(120t)\dfrac{e^{100t}}{100} \\ \\ &= \dfrac{e^{100t}\cos(120t)}{100}+\dfrac{6}{5}\int \sin(120t)e^{100t}\;dt \end{align}

using integration by parts again,

\begin{align*} u&=\sin(120t) & dv&=e^{100t} \\ du&=120 \cos(120t) & v&=\dfrac{e^{100t}}{100} \end{align*}

plugging into $(2)$,

\begin{align*} \int e^{100t} \cos(120t) \;dt &= \dfrac{e^{100t}\cos(120t)}{100}+\dfrac{6}{5} \left[ \dfrac{e^{100t}\sin(120t)}{100}-\int 120\cos(120t) \dfrac{e^{100t}}{100}\;dt\right]\\ \int e^{100t}\cos(120t)\;dt &= \dfrac{e^{100t}\cos(120t)}{100}+\dfrac{6}{5}\dfrac{e^{100t}\sin(120t)}{100}-1.44\int e^{100t}\cos(120t)\;dt \end{align*}

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