Question

The magnitude and direction of currents in different branches are shown in the circuit.

• A. $I=I_L+I_C$. if inductor and cxapacitor are pure.
• B. $I=I_L-I_C$. if inductor and cxapacitor are pure.
• C. $I=I_L+I_C$. if inductor and cxapacitor are not pure.
• D. $I=I_L-I_C$. if inductor and cxapacitor are not pure.

Answer 1

The correct option is C $I=I_L+I_C$. if inductor and cxapacitor are not pure.

$I_C=\left(\frac{V_O}{V_L}\right)\sin \left(\omega t+\frac{\pi }{2}\right)$

$I_L=\left(\frac{V_O}{V_L}\right)\sin \left(\omega t-\frac{\pi }{2}\right)$

$I=I_C+I_L$

Hence,

option $\left(C\right)$ is correct answer.