Answer to Question #142417 in Electrical Engineering for Jisha

Question #142417
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?
1
Expert's answer
2020-11-09T14:19:00-0500

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,


VR=15VV_R = 15V

VL=22VV_L = 22V

VS=48VV_S = 48 V


From the phasor diagram (VCVL)=(VS2VR2)12=(482152)12=45.6V(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=(VCVL)+VL=45.6+22=67.6VVc = (V_C - V_L) +V_L = 45.6+22 = 67.6V

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