Asked by shanza
13. A generator produces 100 kW of power at a potential difference of 10 kV. The power is transmitted through cables of total resistance 5. How much power is dissipated in the cables?
A. 50 W
B. 250 W
C. 500 W
D. 1000 W
E. 50000 W
A. 50 W
B. 250 W
C. 500 W
D. 1000 W
E. 50000 W
Answers
Answered by
Steve
100kW/10kV = 10 A
10A x 5Ω = 50W
10A x 5Ω = 50W
Answered by
shanza
in answer of steve,first point is that i was not understanding. answer is 500 and it is : P=I2R
100*5=500
100*5=500
Answered by
Aiden
P=V×I
=>100 kW=100 kV × I
=>I= 10 A
P=V×I
V=I×R
=>P=I^2×R
P=100×5=500 W
✓✓
WD™
=>100 kW=100 kV × I
=>I= 10 A
P=V×I
V=I×R
=>P=I^2×R
P=100×5=500 W
✓✓
WD™
Answered by
Zeenat
P= 100kW
V=10kW
R=5
power dissipation=?
I=
P=V×I
I=100/10=10A
P=I*2×R
P=10×10×5=500W
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