An RLC circuit with R = 25 5 Ohm, L = 310 mH. and C = 43.1 mu F is connected to

Question

An RLC circuit with R = 25 5 Ohm, L = 310 mH. and C = 43.1 mu F is connected to an ace generator with an ms voltage of 23 V. Determine the average power delivered to this circuit when the frequency of the generator is equal to the resonance frequency. Express your answer using two significant figures. Determine the average power delivered to this circuit when the frequency of the generator is twice the resonance frequency. Express your answer using two significant figures. Determine the average power delivered to this circuit when the frequency of the generator is half the resonance frequency. Express your answer using two significant figures.

Explanation / Answer

As we know that, at the resonance frequency -

Total impedance = R = 25.5 ohm

Again, resonance frequency = f = 1/(2*pi*sqrt(L*C)) = 1/(2*3.14*sqrt(310*10^-3*43.1*10^-6))

= 43.56 Hz.

So, current(rms) = 22/R = 23 / 25.5 = 0.902 A

Part A:

Power = 0.5*Vrms*Irms = 0.5*23*0.902 = 10.37 W

Answer in two significant figures = 10 W

Part B;
Now, frequency = 2 f = 2*43.56 = 87.12 Hz
so inductance= j(2*3.14*87.12*310*10^-3) = j169.6 ohm

And, capacitance = -j /(2*3.14*87.12*43.1*10^-6) = -j42.4 ohm
so net impedance = 25.5 + j127.2 ohms = 129.7 < 78.7 degrees
so current magnitude = 23/129.7 = 0.177 A
And, power = 0.5*23*0.177*cos(78.7) = 0.40 W

Part C:

Again, frequency = f/2 = 43.56/2 = 21.78 Hz

so inductance= j(2*3.14*21.78*310*10^-3) = j 42.4 ohm

And, capacitance = -j /(2*3.14*21.78*43.1*10^-6) = -j169.6 ohm
so net impedance = 25.5 - j127.2 ohms = 129.7 < - 78.7 degrees
so current magnitude = 23/129.7 = 0.177 A
And, power = 0.5*23*0.177*cos(-78.7) = 0.40 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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