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The switch in the circuit shown in Fig. has been closed for a long time. The switch opens at t=0t=0. Find ve(t)v_e(t) for t0t \geq 0.


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At first sight it's possible that we are scared about the ΔY\Delta-Y transformation of the resistors of 70,2070, 20 and 10 Ω10\ \Omega. But it's completely unnecessary. Let's find the initials values of some variables in the circuit.

At t=0t=0^-, CC and LL behave like an open- and short-circuit respectively, so, R20ΩR_{20\Omega} will be in series with R10ΩR_{10\Omega}, and both are in parallel with R70ΩR_{70\Omega}, the equivalent resistance would be:

Req=(R20Ω+R10Ω)R70Ω=21 Ω\begin{align*} R_{eq} &= (R_{20\Omega}+R_{10\Omega})\|R_{70\Omega}= 21 \ \Omega \end{align*}

then, the current in that moment is:

iL(0)=240VR8Ω+R21Ω+R11Ω=6 A\begin{align*} i_{L(0^-)}&=\frac{240\mathrm{V}}{R_{8\Omega}+R_{21\Omega}+R_{11\Omega}}\\ &=6\ \mathrm{A} \end{align*}

Thus, vo(0)v_{o(0^-)} is the voltage across the path containing the voltage source, the R8ΩR_{8\Omega} and the R20ΩR_{20\Omega} resistors. From the mesh of the left side of the source we may calculate the voltage drops in R70ΩR_{70\Omega}, that it's the same voltage that drops across R20ΩR_{20\Omega} and R10ΩR_{10\Omega}. Then we could find the current ixi_x across R20ΩR_{20\Omega}, and then its voltage drops: Making an KVL in the mesh at left side of the source, we obtain:

0=240+V70Ω+iL(0)R11ΩV70Ω=174 V\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:

ix=174V10Ω+20Ω=5.8 A\begin{align*} i_x&=\frac{174\mathrm{V}}{10\Omega+20\Omega}\\ &=5.8\ \mathrm{A} \end{align*}


V20Ω=ixR20Ω=116 V\begin{align*} V_{20\Omega}&=i_x\cdot R_{20\Omega} \\ &=116\ \mathrm{V} \end{align*}

However, we could obtain v20Ωv_{20\Omega} more directly by applying a voltage divisor between R20ΩR_{20\Omega} and R10ΩR_{10\Omega}, namely: V20Ω=174V20Ω10Ω+20Ω=116 VV_{20\Omega}=174\mathrm{V}\dfrac{20\Omega}{10\Omega+20\Omega}=116\ \mathrm{V}.


vo(0)=240iL(0)R8ΩV20Ω=75 V\begin{align*} v_{o(0^-)}&=240-i_{L(0^-)}R_{8\Omega}-V_{20\Omega} \\ &=75\ \mathrm{V} \end{align*}

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