A 13.6- resistor, an 11.9-μF capacitor, and a 19.1-mH inductor are connected in series with a 117-V generator.
(a) At what frequency is the current a maximum?
(b) What is the maximum value of the RMS current?
Note: The ac current and voltage are RMS values and power is an average value unless indicated otherwise.

Explanation:

Given the following information,

Resistor of resistance R = 13.6Ω

Capacitor of capacitance C = 11.9-μF

C = 11.9 × 10^ -6 F

Inductor of inductance L = 19.1-mH

L = 19.1 ×10^-3 H

All this connected in series to a generator that generates Vrms= 117V

Vo = Vrms√2 = 117√2

Vo = 165.463V

a. Frequency for maximum current?

Maximum current occurs at resonance

I.e Xc = XL

At maximum current, the frequency is given as

f = 1/(2π√LC)

Then,

f = 1/(2π√(19.1×10^-3 × 11.9×10^-6)

f = 1/(2π√(2.2729×10^-7))

f = 1/(2π × 4.77 ×10^-4)

f = 333.83Hz

Then, the frequency is 333.83Hz.

b. Since we know the frequency,

Then, we need to find the capacitive and inductive reactance

Capacitive reactance

Xc = 1/2πfC

Xc = 1/(2π × 338.83 × 11.9×10^-6)

Xc = 1/ 0.024961

Xc = 40.1Ω

Also, Inductive reactance

XL = 2πfL

XL = 2π × 333.83 × 19.1×10^-3

XL = 40.1Ω

As expected, Xc=XL, resonance

Then, the impedance in AC circuit is given as

Z = √ (R² + (Xc—XL)²)

Z = √ 13.6² + (40.1-40.1)²)

Z = √13.6²

Z = 13.6 ohms

Then, using ohms las

V = IZ

Then, I = Vo/Z

Io = 165.46/13.6

Io = 12.17Amps

The current is 12.17 A

answer; equivalent to a one(1) horsepower motor;