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Question

# The switch in the circuit shown in Fig. has been closed for a long time. The switch opens at $t=0$. Find $v_e(t)$ for $t \geq 0$.

Solution

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At first sight it's possible that we are scared about the $\Delta-Y$ transformation of the resistors of $70, 20$ and $10\ \Omega$. But it's completely unnecessary. Let's find the initials values of some variables in the circuit.

At $t=0^-$, $C$ and $L$ behave like an open- and short-circuit respectively, so, $R_{20\Omega}$ will be in series with $R_{10\Omega}$, and both are in parallel with $R_{70\Omega}$, the equivalent resistance would be:

\begin{align*} R_{eq} &= (R_{20\Omega}+R_{10\Omega})\|R_{70\Omega}= 21 \ \Omega \end{align*}

then, the current in that moment is:

\begin{align*} i_{L(0^-)}&=\frac{240\mathrm{V}}{R_{8\Omega}+R_{21\Omega}+R_{11\Omega}}\\ &=6\ \mathrm{A} \end{align*}

Thus, $v_{o(0^-)}$ is the voltage across the path containing the voltage source, the $R_{8\Omega}$ and the $R_{20\Omega}$ resistors. From the mesh of the left side of the source we may calculate the voltage drops in $R_{70\Omega}$, that it's the same voltage that drops across $R_{20\Omega}$ and $R_{10\Omega}$. Then we could find the current $i_x$ across $R_{20\Omega}$, and then its voltage drops: Making an KVL in the mesh at left side of the source, we obtain:

\begin{align*} 0&=-240+V_{70\Omega}+i_{L(0^-)}\cdot R_{11 \Omega}\\ V_{70\Omega}&=174 \ \mathrm{V} \end{align*}

then, if we applying Ohm's law:

\begin{align*} i_x&=\frac{174\mathrm{V}}{10\Omega+20\Omega}\\ &=5.8\ \mathrm{A} \end{align*}

and

\begin{align*} V_{20\Omega}&=i_x\cdot R_{20\Omega} \\ &=116\ \mathrm{V} \end{align*}

However, we could obtain $v_{20\Omega}$ more directly by applying a voltage divisor between $R_{20\Omega}$ and $R_{10\Omega}$, namely: $V_{20\Omega}=174\mathrm{V}\dfrac{20\Omega}{10\Omega+20\Omega}=116\ \mathrm{V}$.

finally

\begin{align*} v_{o(0^-)}&=240-i_{L(0^-)}R_{8\Omega}-V_{20\Omega} \\ &=75\ \mathrm{V} \end{align*}

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