Respuesta :
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
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