A 100-$\varOmega$ resistor,an 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 inductance of the inductor?
A)20.0 mH
B)40.0 mH
C)60.0 mH
D)80.0 mH

Question

Quizplus · Accepted Answer

We can use the formula for impedance of series RL circuit:
Impedance Z = sqrt(R^2 + (XL - XC)^2)

Given Z = 339 ohms, R = 100 ohms, XC = 1/(2*pi*f*C) = 530.93 ohms, f = 1000 Hz, V = 10 V

Substituting these values in the formula, we get:
339 = sqrt(100^2 + (XL - 530.93)^2)
Squaring both sides, we get:
115,321 = 10,000 + (XL - 530.93)^2
105,321 = (XL - 530.93)^2
Taking square root on both sides, we get:
XL - 530.93 = +/-325.005
XL = 855.935 ohms or 205.925 ohms

We can discard the smaller value since it doesn't make physical sense for the inductance to be that small. Therefore, the inductance of the inductor is:
XL = 855.935 ohms = 2*pi*f*L
L = XL/(2*pi*f) = 0.060 H = 60 mH, which is closest to option C.

A 100- Ω\varOmegaΩ Resistor,an Inductor and a 3 Ω\varOmegaΩ What Is the Inductance of the Inductor? A)20

A 100- $\varOmega$ Resistor,an Inductor and a 3 $\varOmega$ What Is the Inductance of the Inductor?
A)20