HW 10 LRC Circuits Problem 11 Constants Part A In an L-R-C series cirouit, the r

Question

HW 10 LRC Circuits Problem 11 Constants Part A In an L-R-C series cirouit, the resistance is 340 ohms, the nductance is 0.400 henrys, and the capacitance is 2.00 102 microfarads What s the resonance angular frequency uwp of the cirout? Express your answer in radians per second to three significant figures View Available Hints) rad/s Submit Part The capacitor can withstand a peak voitage of 560 volts If the voltage source operates at the resonance fregueney what masimum volage s can the wource have ithe maxmum capacitor voltage is not excooded Express your answer in voits to three significant figures View Available Hints) Submit Provide Feedback

Explanation / Answer

L = 0.400 H

C = 2.00 x 10^-2 micro F = 2.00 x 10^-2 x 10^-6 = 2.00 x 10^-8 F

R = 340 ohm

Part A -

Resonance angular frequency w0 = 1 / sqrt(LC) = 1 / sqrt(0.40 x 2.00 x 10^-8) = 10^4 / 0.8944 = 1.12 x 10^4 rad/s

Part B -

Now we have the expression -

Capacitor voltage at resonance = current*capacitor impedance at resonance.

= (supply volts/resistance)*(1/w0*c)

= (supply voltage/340)*(1/1.12*10^4*2.0*10^-8) = 560 to avoid breakdown

=> Supply volts = 340*560*1.12*10^4*2.0*10^-8

= 42.65 V

Therefore, the requisite maximum voltage, Vmax = 42.65 V

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.