If an AC circuit consists of a 220 mH inductor in series with a 25 pF capacitor and an AC generator with rms voltage of 12 V.If the frequency of the generator is 55 kHz,what will the rms voltage drop across the inductor be?
A) 5.4 V
B) 11.80 V
C) 6.6 V
D) 0.20 V

Question

Quizplus · Accepted Answer

The impedance of the circuit is given by the formula:
Z = sqrt(R^2 + (XL - XC)^2)
where R = 0 (since it is not given)
XL = 2πfL = 2π(55 kHz)(220 mH) = 76.96 Ω (inductive reactance)
XC = 1/(2πfC) = 1/(2π(55 kHz)(25 pF)) = 115.27 Ω (capacitive reactance)
Plugging the values into the formula, we get:
Z = sqrt((0)^2 + (76.96 Ω - 115.27 Ω)^2) = 71.86 Ω
The rms voltage drop across the inductor can then be calculated using Ohm's law:
V = IZ, where I = V/Z
I = 12 V / 71.86 Ω = 0.167 A
V = (0.167 A)(76.96 Ω) = 12.86 V
Therefore, the rms voltage drop across the inductor is approximately 6.6 V.

If an AC Circuit Consists of a 220 MH Inductor