Question

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?

Answer 1

Answer:

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

Answer 2

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