A series AC circuit is shown.The inductor has a reactance of 50 Ω and an inductance of 250 mH.A 40 Ω resistor and a capacitor whose reactance is 150 ohms are also in the circuit.The rms current in the circuit is 2.0 A.In the figure,the rms voltage of the source is closest to:

Question

Quizplus · Accepted Answer

Using Kirchhoff's circuit laws, we can write:
VR + VL + VC = VS
where VR, VL, VC are the voltage drops across the resistor, inductor, and capacitor respectively, and VS is the voltage of the source.

The rms current in the circuit is given as:
Irms = 2.0 A

The reactance of the inductor is given as:
XL = 50 Ω

The inductance of the inductor is given as:
L = 250 mH = 0.25 H

The reactance of the capacitor is given as:
XC = 150 Ω

Using these values, we can calculate the impedance of the circuit as:
Z = sqrt(R^2 + (XL - XC)^2)
   = sqrt(40^2 + (50 - 150)^2)
   = sqrt(40^2 + (-100)^2)
   = 107.7 Ω

The rms voltage of the source can be calculated as:
VS = Irms * Z
   = 2.0 * 107.7
   = 215.4 V

Therefore, the closest option is A) 220 V.

A Series AC Circuit Is Shown