Q21GATE 2004MCQ

In figure, the admittance values of the elements in Siemens are $Y_{R}$ = 0.5 + j0, $Y_{L}$ = 0 - j1.5, $Y_{C}$ = 0 + j0.3 respectively. The value of I as a phasor when the voltage E across the elements is 10 $\angle 0^{\circ}$ V is

🚀 Solve in Practice Mode

📖 Explanation

$I_R=Y_RE$ $\;\;=(0.5+j0) \times 10\angle 0^{\circ}=5A$ $I_Y=Y_LE$ $\;\;=(0-j1.5)\times 10\angle 0^{\circ}=-j15A$ $I_C=Y_CE$ $\;\;=(0+j0.3)\times 10\angle 0^{\circ}=j3A$ $I=I_R+I_Y+I_C$ $\;\;=5+(-j15)+j3 \;\;=5-j12 A$