Question

For the circuit given below, the current I in the circuit is ________

(a) –j1 A

(b) J1 A

(c) Zero

(d) 20 A

Answer 1

Correct answer is (a) –j1 A

The best I can explain: XEQ = sL + \(\frac{R×1/sC}{R+1/sC} = sL + \frac{R}{1+sRC}\)

IO = \(\frac{V}{X_{EQ}}\)

∴ I = \(\frac{X_C}{X_C+R}\) IO

= \(\frac{1/sC}{\frac{1}{sC}+R} × \frac{V}{\frac{sL(1+sRC)+R}{(1+sRC)}}\)

= \(\frac{1}{1+sRC} × \frac{V}{\frac{sL(1+sRC)+R}{(1+sRC)}}\)

= \(\frac{V}{sL(1+sRC)+R}\)

= \(\frac{V}{j×10^3×20×10^{-3} (1+j×10^3×50×10^{-6}+1)}\)

= \(\frac{V}{20j(1+j50×10^{-3})+1}\)

= \(\frac{V}{20j-1+1} = \frac{20}{20j}\) = -j1 A.