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## Homework Statement

http://img159.imageshack.us/img159/1867/es100wi8.th.png [Broken]http://g.imageshack.us/thpix.php [Broken]

a) Write an expression for V1 and V2 both in time domain and phasor domain (Solved)

b) Write an expression for the current I both in time domain and phasor domain (Solved)

c) Calculate the Power factor of the supply and specify whether it is lagging or

leading (Solved)

**d) Specify the type of the reactance (XC or XL)**

e) Determine the value of X and hence the corresponding value of L or C

f) Calculate the supply average, reactive and apparent power

g) Draw the phasor diagram of the circuit

e) Determine the value of X and hence the corresponding value of L or C

f) Calculate the supply average, reactive and apparent power

g) Draw the phasor diagram of the circuit

## Homework Equations

f= 250/3

## The Attempt at a Solution

a)

V1 = 10sqrt(2)<48 = 9.46+10.51

V2 = 5/sqrt(2)<0 = 3.54

b)

I = v2/r2 = 0.354<0

c)

PF = 0.669, lagging

d)

X is an inductor

e)

VCh2 + Vxl = Vsource

Vxl = 9.46+10.51j - 5/sqrt(2) = 5.93 +10.51j = 12.1<60.6

Zx = Vxl/Is = 12.1<60.6 / 0.354<0 = 16.77+ 29.73j

This tells me that there's a physical impedance associated with the coil, equal in value to 16.77ohms...Meaning that it's not a pure inductor?

To find L:

29.73 = 2*pi*f*L ... 29.73/ (2*pi*250/3) = L = 0.057H

But these results don't add up to the next few parts:

-Active Power-

P(10ohm)=I^2*R=(0.354)^2*10 = 1.25W

P(16.77ohm)=I^2*R=(0.354)^2*16.77 = 2.10W

Total:3.35W

-Reactive Power-

P(XL)=I^2*R=(0.354)^2*29.73 = 3.72VAR

-Apparent Power-

Active+Reactive*j = 3.35+3.72j

which matches:

S=EI=10sqrt(2)<48 * 0.354<0 = 3.35 + 3.72j

Have I done this question correctly? Is it possible that the inductor has a simple resistance component to it?

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