Question

A 1.5-kΩ resistor and 30-mH inductor are connected
in series, as shown below, across a 120-V (rms) ac power
source oscillating at 60-Hz frequency. (a) Find the current
in the circuit. (b) Find the voltage drops across the resistor
and inductor. (c) Find the impedance of the circuit. (d) Find
the power dissipated in the resistor. (e) Find the power
dissipated in the inductor. (f) Find the power produced by
the source.

Answer 1

Solution ) R =1.5 Kohms =1.5×1000=1500 ohms

L= 30mH = 30×10^(-3) H

Vrms = 120V

f=60Hz

(a) current I =?

I = (Vrms)/(Z)

Z is impedance

Z = ( R^(2) + (XL)^2 )^(1/2)

XL = WL = (2(pi)f)L= 2×3.14×60×30×10^(-3) = 11.304

Z = ( 1500^2 + 11.304^2 )^(1/2)

Z = 1500.04 ohms

I = 120/1500.04

I = 0.08 A

(b) voltage drop across resistor Vr = IR

Vr = 0.08×1500=120

No voltage drop occurs across inductor

(C) impedance Z = 1500.04 ohms from PART (a)

(d) power dissipated in resistor Pr = (I^2)R

Pr =( (0.08)^2)×1500

Pr =9.6 W

(e) no power dissipated in inductor

(f) Power P = (Vrms)(Irms)

P=120×0.08 =9.6W