In an L-R-C series circuit, L =0.280 H and C =4.00 F. The voltage amplitude of t

Question

In an L-R-C series circuit, L=0.280 H and C=4.00 F. The voltage amplitude of the source is 120 V.

1-What is the resonance angular frequency of the circuit?

2-When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70

A. What is the resistance R of the resistor?

3-

At the resonance angular frequency, what are the peak voltages across the inductor?

4-

At the resonance angular frequency, what are the peak voltages across the capacitor?

5-At the resonance angular frequency, what are the peak voltages across the resistor?

Explanation / Answer

1.

at resonance angular frequency f0 ,

Xc = XL

1 / 2 pi f0 C   = 2pi f0 L

f0 = 150.39 Hz

2. at resonance frequency , impedance of circuit Z = R

so I = V/R

1.70 = 120 / R

R = 70.59 ohm

3. Xc = XL = 2pifL = 2 x pi x 150.39 x 0.280 = 264.58 ohm

peak voltage across inductor = i XL = 1.70 x 264.58 =450 V

4. it will be same = 450 V

5. across resistor = 120 V

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