Question

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

A) what is the resonance angular frequency of the circuit?
B) 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?
C) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Answer 1

The resonance angular frequency of the L-R-C series circuit is 645.93 rad/s. The resistance of the resistor is 70.588 Ω. At resonance, the peak voltages across the inductor, capacitor, and resistor are equal.

Step-by-step explanation:

A) The resonance angular frequency of an L-R-C series circuit can be calculated using the formula:

ωr = 1 / √(LC)

Substituting the given values, we have:

ωr = 1 / √(0.280 H × 4.00 μF)

ωr = 645.93 rad/s

B) At resonance, the current amplitude in the circuit can be calculated using the formula:

Ia = Vs / R

Substituting the given values, we have:

1.70 A = 120 V / R

Solving for R, we get:

R = 70.588 Ω

C) At resonance, the peak voltages across the inductor, the capacitor, and the resistor are equal.

Answer 2

The resonance angular frequency of the L-R-C series circuit is 39.283 rad/s. The resistance R of the resistor is 70.588 ohms. At resonance, the peak voltages across the inductor, capacitor, and resistor are all 120 V.

Step-by-step explanation:

A) The resonance angular frequency of the circuit can be calculated using the formula:

ω = 1/(√LC)

Substituting the given values, ω = 1/(√(0.280 H * 4.00 µF)) = 39.283 rad/s

B) The current amplitude in the circuit at resonance can be calculated using the formula:

I = V/R

Substituting the given values and solving for R, R = V/I = 120 V / 1.70 A = 70.588 ohms

C) At resonance, the voltages across the inductor, capacitor, and resistor are equal. Therefore, the peak voltage across each component is equal to the voltage amplitude of the source, which is 120 V.