The circuit shown in Fig. 7.10 a has reached steady state with the switch in position 1. At time t = 0 the switch moves from position 1 to position 2 . We want to calculate vo(t) for t 0.

Question

The circuit shown in Fig. 7.10 a has reached steady state with the switch in position 1. At time t = 0 the switch moves from position 1 to position 2 . We want to calculate [latex]v_o[/latex](t) for t > 0.

Accepted Answer

Step 1. [latex]v_{o}(t)[/latex] is of the form [latex]K_{1}+K_{2} e^{-t / au}[/latex].

Step 2. Using the circuit in Fig. 7.10 b, we can calculate [latex]i_{L}\left(0- ight)[/latex]

[latex]i_{A}=\frac{12}{4}=3 \mathrm{~A}[/latex]

Then

[latex]i_{L}(0-)=\frac{12+2 i_{A}}{6}=\frac{18}{6}=3 \mathrm{~A}[/latex]

Step 3. The new circuit, valid only for t = 0+, is shown in Fig. 7.10c. The value of the current source that replaces the inductor is [latex]i_{L}(0-)=i_{L}(0+)=3 \mathrm{~A}[/latex]. Because of the current source

[latex]v_{o}(0+)=(3)(6)=18 \mathrm{~V}[/latex]

Step 4. The equivalent circuit, for the steady-state condition after switch closure, is given in Fig. 7.10d. Using the voltages and currents defined in the figure, we can compute [latex]v_{o}(\infty)[/latex] in a variety of ways. For example, using node equations we can find [latex]v_{o}(\infty)[/latex] from

[latex]\frac{v_{B}-36}{2}+\frac{v_{B}}{4}+\frac{v_{B}+2 i_{A}^{\prime}}{6} =0 [/latex]

[latex]i_{A}^{\prime} =\frac{v_{B}}{4} [/latex]

[latex]v_{o}(\infty) =v_{B}+2 i_{A}^{\prime}[/latex]

or, using loop equations,

[latex]36 =2\left(i_{1}+i_{2} ight)+4 i_{1} [/latex]

[latex]36 =2\left(i_{1}+i_{2} ight)+6 i_{2}-2 i_{1} [/latex]

[latex]v_{o}(\infty) =6 i_{2}[/latex]

Using either approach, we find that [latex]v_{o}(\infty)=27 \mathrm{~V}[/latex].

Step 5. The Thévenin equivalent resistance can be obtained via [latex]v_{\mathrm{oc}}[/latex] and [latex]i_{\mathrm{sc}}[/latex] because of the presence of the dependent source. From Fig. 7.10 e we note that

[latex]i_{A}^{\prime \prime}=\frac{36}{2+4}=6 \mathrm{~A}[/latex]

Therefore,

[latex]v_{\mathrm{oc}} =(4)(6)+2(6) [/latex]

[latex]=36 \mathrm{~V}[/latex]

From Fig. 7.10 f we can write the following loop equations:

[latex]36=2\left(i_{A}^{\prime \prime \prime}+i_{\mathrm{sc}} ight)+4 i_{A}^{\prime \prime \prime} [/latex]

[latex]36=2\left(i_{A}^{\prime \prime}+i_{\mathrm{sc}} ight)+6 i_{\mathrm{se}}-2 i_{A}^{\prime \prime \prime}[/latex]

Solving these equations for [latex]i_{\mathrm{se}}[/latex] yields

[latex]i_{\mathrm{sc}}=\frac{9}{2} \mathrm{~A}[/latex]

Therefore,

[latex]R_{\mathrm{Th}}=\frac{v_{\mathrm{oc}}}{i_{\mathrm{sc}}}=\frac{36}{9 / 2}=8 \Omega[/latex]

Hence, the circuit time constant is

[latex] au=\frac{L}{R_{\mathrm{Th}}}=\frac{3}{8} \mathrm{~s}[/latex]

Step 6. Using the information just computed, we can derive the final equation for [latex]v_{o}(t)[/latex] :

[latex]K_{1}=v_{o}(\infty)=27 [/latex]

[latex]K_{2}=v_{o}(0+)-v_{o}(\infty)=18-27=-9[/latex]

Therefore,

[latex]v_{o}(t)=27-9 e^{-t /(3 / 8)} \mathrm{V}[/latex]

Question 7.5: The circuit shown in Fig. 7.10 a has reached steady state wi...

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