Question

When a series RLC circuit is subject to 48 V, VR is 15 V, and VL is 22 V. What is the voltage
across the capacitor?

Answer 1

Base on the data given in the question a circuit could be constructed as shown,

The phasor diagram for a RLC Series circuit could be drawn as follows,

Comparing the phasor diagram and given information following datas are given,

$V_R = 15V$

$V_L = 22V$

$V_S = 48 V$

From the phasor diagram $(V_C - V_L) = {(V_S^2 - V_R^2)}^\frac{1}{2} = (48^2-15^2)^\frac{1}{2} = 45.6V$

There fore $Vc = (V_C - V_L) +V_L = 45.6+22 = 67.6V$