question_answer

In the circuit diagram shown, \[{{X}_{C}}=100\Omega ,\]\[{{X}_{L}}=200\Omega \] & \[R=100\Omega .\] The effective current through the source is -

A) 2A

B) \[2\sqrt{2}A\]

C) 0.5 A

D) \[\sqrt{0.4}\,A\]

Correct Answer: B

Solution :

\[{{I}_{R}}=\frac{V}{R}=\frac{200}{100}=2A\]

\[I'=\frac{V}{{{X}_{L}}-{{X}_{C}}}=\frac{200}{100}=2A\]

\[I=\sqrt{I_{R}^{2}+I{{'}^{2}}}=2\sqrt{2}\,\,Amp.\]