Transcript Example 10.5 1 ECE 201 Circuit Theory 1

Example 10.5
ECE 201 Circuit Theory 1
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A load having an impedance of 39 + j26 Ω is fed from a voltage source
through a line having an impedance of 1 + j4 Ω. The effective, or rms,
value of the source voltage is 250 V.
Calculate the load current IL and the voltage VL.
ECE 201 Circuit Theory 1
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Calculate the load current IL and the voltage VL.
The line and load impedances are in series across the voltage source,
so the load current equals the voltage divided by the total impedance.
The load voltage equals the load current multiplied by the load
impedance.
2500
2500

 5  36.87 A
40  j 30 5036.87
VL  I L Z L  (39  j 26) I L  (46.8733.69)(5  36.87)
IL 
VL  234.35  3.18V
ECE 201 Circuit Theory 1
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Calculate the active and reactive power delivered to the load.
S  VL I L*  (234  j13)( 4  j 3)
S  (234.36  3.18)(536.87)
S  1171.833.69
S  975  j 650VA
Average Power = 975 W
Reactive Power = 650 VARS
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Calculate the average and reactive power delivered to the line.
2
P  I eff R
P  (5) 2 (1)  25W
2
Q  I eff X
Q  (5) 2 (4)  100VAR
Reactive Power is positive due to
an inductive line reactance
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Calculate the average and reactive power supplied by the source
Add complex powers delivered to the line and load
S  25  j100  975  j 650
S  1000  j 750VA
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*
*
S  Veff I eff
 Veff I eff
Calculate the apparent power
*
*
SS  Veff I eff
 Veff I eff
The – sign is used whenever the current reference is in the
direction of a voltage rise
S S  250(4  j3)  (1000  j 750)VA
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