Question
In the given circuit, with steady current, calculate the potential drop across the capacitor in terms of V.
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Answer
In steady state branch BE is eliminated
$ \text{I}=\frac{10{\text{V}}-5\text{V}}{(3+2){\Omega}}\text{A}$
$=\text{1A}$
For loop EBCDE
$-v_c{-5+10-3\times1}=0$
$-{\text{V}}_c+10-8=0$
$\therefore{\text{V}_c}=2\text{ }\text{volt}$
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