Example 10.5 1ECE 201 Circuit Theory 1. A load having an impedance of 39 + j26 Ω is fed from a...
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Transcript of Example 10.5 1ECE 201 Circuit Theory 1. A load having an impedance of 39 + j26 Ω is fed from a...
Example 10.5
1ECE 201 Circuit Theory 1
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.
2ECE 201 Circuit Theory 1
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.
VV
IjZIV
Aj
I
L
LLLL
L
18.335.234
)87.365)(69.3387.46()2639(
87.36587.3650
0250
3040
0250
3ECE 201 Circuit Theory 1
Calculate the active and reactive power delivered to the load.
VAjS
S
S
jjIVS LL
650975
69.338.1171
)87.365)(18.336.234(
)34)(13234(*
Average Power = 975 W Reactive Power = 650 VARS
4ECE 201 Circuit Theory 1
Calculate the average and reactive power delivered to the line.
VARQ
XIQ
WP
RIP
eff
eff
100)4()5(
25)1()5(
2
2
2
2
Reactive Power is positive due to an inductive line reactance
5ECE 201 Circuit Theory 1
Calculate the average and reactive power supplied by the source
Add complex powers delivered to the line and load
VAjS
jjS
7501000
65097510025
6ECE 201 Circuit Theory 1
Calculate the apparent power**effeffeffeff IVIVS
The – sign is used whenever the current reference is in the direction of a voltage rise
VAjjSS )7501000()34(250
**effeffeffeffS IVIVS
7ECE 201 Circuit Theory 1