A 25.0-mH inductor, a 2.00-μF capacitor, and a certain resistor are connected in series across an ac voltage source at 1000 Hz. If the impedance of this circuit is 200 Ω, what is the value of the resistor?
A) 100 Ω
B) 184 Ω
C) 552 Ω
D) 579 Ω
E) 200 Ω

Question

Quizplus · Accepted Answer

At resonance, the reactance of the inductor and the capacitor will cancel each other out, leaving only the resistance in the circuit. We can use the formula for impedance of a series RLC circuit at resonance to find the value of the resistor:

Z = R

Z = √(R^2 + (XL - XC)^2)  (at resonance, XL = XC)

Z = √(R^2 + 0) = R

200 Ω = R

Therefore, the value of the resistor is 200 Ω.

A 250-MH Inductor, a 2