An AC source with $\Delta$V_max = 220 V and f = 60.0 Hz is connected to an RLC series combination.The resistance,R,inductance,L and capacitance,C of this circuit are 80.0 $\varOmega$,0.300 H,and 0.02 mF,respectively.The maximum current through the circuit is A) 5.98 A. B) 2.75 A. C) 1.44 A. D) 2.67 A. E) 5.56 A.

Question

Quizplus · Accepted Answer

First, we need to find the resonant frequency of the circuit using the formula:

fr = 1 / 2π√(LC)

Plugging in the values, we get:

fr = 1 / 2π√(0.300 H x 0.02 mF)
   = 530.93 Hz

Since the applied frequency (60 Hz) is much lower than the resonant frequency, we can assume that the circuit is operating in the underdamped regime.

The impedance of the circuit at resonance is:

Z = R

So the maximum current through the circuit can be found using Ohm's Law:

I(max) = V(max) / R

where V(max) is the maximum voltage of the AC source, which is 220 V.

Therefore:

I(max) = 220 V / 80.0 Ω
       = 2.75 A

Hence, the best choice is D) 2.67 A, which is the closest to the calculated value of 2.75 A.

An AC Source With Δ\DeltaΔ Vmax = 220 V and F = 60

An AC Source With $\Delta$ V_max = 220 V and F = 60