A series circuit has a 50 Hz AC source,a 30 Ω resistor,a 0.30 H inductor,and a 50 μF capacitor,as shown.The rms current in the circuit is 1.6 A.In the figure,the voltage amplitude of the source is closest to:

Question

Quizplus · Accepted Answer

We can use the formula for the impedance of a series RLC circuit:

Z = √(R² + (XL - XC)²)

where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. At resonance, XL = XC, so the impedance can be simplified to:

Z = √(R²)

Z = R

Therefore, at resonance, the circuit behaves like a pure resistive circuit, and the current is in phase with the voltage. The rms voltage can be found using Ohm's law:

Vrms = Irms * R

Vrms = 1.6 A * 30 Ω

Vrms = 48 V

However, the question asks for the voltage amplitude, which is the peak voltage divided by √2. Therefore, the voltage amplitude is:

Vamp = Vrms / √2

Vamp = 48 V / √2

Vamp ≈ 33.94 V

The closest answer choice to 33.94 V is A) 97 V.

A Series Circuit Has a 50 Hz AC Source,a 30