In a series LC circuit, VL = 8.3 V and VC = 10.6 V. VS = .
A) -2.3 V 2-90°
B) 8.3 V 2-90°
C) -2.3 V 290°
D) 2.3 V 2-90°

Question

Quizplus · Accepted Answer

In a series LC circuit, the source voltage $V_S$ is the difference between the inductor voltage ($V_L$) and the capacitor voltage ($V_C$) because they are 180 degrees out of phase. Here, $V_L = 8.3V$ and $V_C = 10.6V$, so the difference $V_S = V_L - V_C = 8.3V - 10.6V = -2.3V$. The negative sign indicates that $V_S$ is 180 degrees out of phase with $V_C$, but since we're dealing with magnitudes and directions in phasor form, the correct representation is $2.3V$ at an angle of $-90°$ to align with the conventional phasor representation, making option D correct.

In a Series LC Circuit, VL = 8