Consider the following circuit, where it is given than vC(0) = 10V. • Find the expression for the capacitor voltage vC(t) as a function of time for t 0. • Find the expression for the capacitor current iC(t) as a function of time for t 0. • Find the energy stored in the capacitor at t = 5 ln (2)

Question

Consider the following circuit, where it is given than [latex]v_{C}(0)=10 \mathrm{~V}[/latex].

Find the expression for the capacitor voltage [latex]v_{C}(t)[/latex] as a function of time for [latex]t>0[/latex].
Find the expression for the capacitor current [latex]i_{C}(t)[/latex] as a function of time for [latex]t>0[/latex].
Find the energy stored in the capacitor at [latex]t=5 \ln (2) \mathrm{s}[/latex].

Accepted Answer

For [latex]v_{\mathrm{oc}}[/latex] seen by the capacitor:

[latex]v_{x}=3 \mathrm{~V}[/latex]
[latex]\operatorname{KVL}(a): 14+i_{a}+2 v_{x}+\left(i_{a}-i_{b}\right)=0 \longrightarrow 2 i_{a}-i_{b}=-20[/latex]
[latex]\operatorname{KVL}(b):-16+\left(i_{b}-i_{a}\right)+(5 / 2) i_{b}=0 \longrightarrow-2 i_{a}+7 i_{b}=32[/latex]
Then [latex]i_{a}=-9 \mathrm{~A}, i_{b}=2 \mathrm{~A}[/latex], and [latex]v_{\mathrm{oc}}=(5 / 2) i_{b}=5 \mathrm{~V}[/latex]

For [latex]i_{\mathrm{sc}}[/latex] seen by the capacitor: [latex]v_{x}[/latex] and [latex]\operatorname{KVL}(a)[/latex] are the same as above.

[latex]\operatorname{KVL}(b):-16+\left(i_{b}-i_{a}\right)=0 \longrightarrow-2 i_{a}+2 i_{b}=32[/latex]
Then [latex]i_{a}=-4 \mathrm{~A}, i_{b}=12 \mathrm{~A}[/latex], and [latex]i_{\mathrm{sc}}=i_{b}=12 \mathrm{~A}[/latex]
And [latex]R_{\mathrm{th}}=v_{\mathrm{oc}} / i_{\mathrm{sc}}=5 / 12 \Omega[/latex]

Equivalent circuit:

[latex]v_{C}(t)=5 \exp (-t / 5)+5 \mathrm{~V}[/latex]
[latex]i_{C}(t)=C \frac{d v_{C}(t)}{d t}=-12 \exp (-t / 5) \mathrm{A}[/latex]
[latex]t=5 \ln (2) \mathrm{s} \longrightarrow v_{C}(t)=5 \exp (-\ln (2))+5=15 / 2 \mathrm{~V}[/latex]

[latex]\longrightarrow w_{C}(t)=675 / 2 \mathrm{~W}[/latex]

Question 6.E.97: Consider the following circuit, where it is given than vC(0)...