A resistor,a 20 mH inductor and a 3.0-µF capacitor are connected in series with a time varying emf of 10.0 V and frequency 1000 Hz.The impedance of the circuit is 339 $\varOmega$ .What is the resistance of the resistor?
A)331 $\varOmega$
B)100 $\varOmega$
C)258 $\varOmega$
D)179 $\varOmega$

Question

Quizplus · Accepted Answer

The impedance of the circuit is given by:

Z = √(R^2 + (ωL - 1/(ωC))^2)

where R is the resistance, L is the inductance, C is the capacitance, and ω is the angular frequency (equal to 2πf, where f is the frequency).

Substituting the given values, we get:

339 = √(R^2 + (2π(1000)(20x10^-3) - 1/(2π(1000)(3x10^-6)))^2)

Solving for R, we get:

R = 331 Ω.

Therefore, the answer is choice letter A.

A Resistor,a 20 MH Inductor and a 3 Ω\varOmegaΩ What Is the Resistance of the Resistor? A)331

A Resistor,a 20 MH Inductor and a 3 $\varOmega$ What Is the Resistance of the Resistor?
A)331