Question

An RLRL circuit with a 5Ω5-\Omega resistor and a 0.050.05-H inductor carries a current of 11 A at t=0,t = 0, at which time a voltage source E(t)=5cos120tVE ( t) = 5 \cos 120t V is added. Determine the subsequent inductor current and voltage.

Solution

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Answered 3 years ago
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Given:\textbf{Given:}

R=5  Ω,L=0.05  H,I(t)=I(0)=1,andE(t)=5cos(120t)  VR=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 subsequent inductor current\text{\underline{subsequent inductor current}}, where,

I(t)=eRt/L[eRt/LE(t)L  dt+K]I(t)=e5t/0.05[e5t/0.055cos(120t)0.05  dt+K]I(t)=e100t[100e100tcos(120t)  dt+K]\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)(1), we will use integration by parts, where,

e100tcos(120t)  dt\begin{align*} \int e^{100t} \cos(120t) \;dt \end{align*}

u=cos(120t)dv=e100tdu=120sin(120t)v=e100t100\begin{align*} u&=\cos(120t) & dv&=e^{100t} \\ du&=-120 \sin(120t) & v&=\dfrac{e^{100t}}{100} \end{align*}

and so,

e100tcos(120t)  dt=e100tcos(120t)100+120sin(120t)e100t100=e100tcos(120t)100+65sin(120t)e100t  dt\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,

u=sin(120t)dv=e100tdu=120cos(120t)v=e100t100\begin{align*} u&=\sin(120t) & dv&=e^{100t} \\ du&=120 \cos(120t) & v&=\dfrac{e^{100t}}{100} \end{align*}

plugging into (2)(2),

e100tcos(120t)  dt=e100tcos(120t)100+65[e100tsin(120t)100120cos(120t)e100t100  dt]e100tcos(120t)  dt=e100tcos(120t)100+65e100tsin(120t)1001.44e100tcos(120t)  dt\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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