The capacitor can withstand a peak voltage of 600 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude VmaxVmaxV_max can the source have if the maximum capacitor voltage is not exceeded?

426.5 V

Explanation:

The complete question is:

In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 henrys, and the capacitance is 1.20*10-2 microfarads.

The capacitor can withstand a peak voltage of 600 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude V_max can the source have if the maximum capacitor voltage is not exceeded?

at resonance

XL = Xc

2πfL=1/2πfC

2*π*f*0.38 = 1 / (2π*f*0.0000012)

f = 235.8 Hz

XL = 2π*235.8*0.38 = 562.7

Xc = 1 / (2π*235.8*0.0000012) = 562.7

at resonance

Z = 400

Vc = - j562.7 / 400 * Vs

600 = - j562.7/400 * Vs

Vs = 426.5V

The question is not complete and the first part of the question says;

In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 henrys, and the capacitance is 1.20*10^ (-2) microfarads.

Answer:

Vs = 42.65 V

Explanation:

Formula for resonance is given as;

f = 1 / (2π (√LC))

Where L is inductance in henrys

While C is Capacitance in farads

In this question;

C = 1.20*10^ (-2) microfarads = 1.2 x 10^ (-8) Farads

L = 0.38 H

Thus,

f = 1 / (2π (√1.20*10^ (-8) x 0.38))

f = 1/0.00042428951

f = 2356.88 Hz

Now, capacitive resistance Xc is given as;

Xc = 1 / (2πfC)

Xc = 1 / (2π x 2356.88 x 1.2 x 10^ (-8)) = 5627.32 ohms

Since the capacitors can withstand a peak voltage of 600V.

Thus Vc = IXc

Where I is current thus, Vc/Xc = I

Also, for the series, Vs/R = I

Thus, Vc/Xc = Vs/R

So, Vc = 600 V, Xc = 5627.32 ohms while R = 400 ohms

Thus,

Making Vs the subject,

(Vc/Xc) x R = Vs

Thus,

(600/5627.32) x 400 = Vs

Vs = 42.65 V