Misconceptions in Calculating Power Dissipation in Electrical Circuits

AI Thread Summary
The discussion centers on calculating power dissipation in electrical circuits, specifically regarding a 100 kW generator supplying power at 10 kV through cables with a resistance of 5 Ohms. The correct supply current is determined to be 10 A using the formula P = IV. A misconception arises when attempting to calculate power dissipation in the cables; using P = (V^2)/R yields an incorrect value due to not accounting for the generator's power limitation. The correct approach is to use P = I^2R, which aligns with the actual current flowing through the circuit. This highlights the importance of understanding the relationship between power, current, and resistance in circuit calculations.
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Homework Statement


A 100 kW generator supplies power at a p.d of 10kV through cables of total resistance 5 Ohms. Calculate the supply current and the power dissipated in the cables.

Homework Equations


P = IV
V=IR
Q=It

The Attempt at a Solution


Supply current:
P=IV
100 kW = (I)(10kV)
I=10A
Supply current 10 A.

But I get different answers when I try to find the power dissipated in the cable.
By using this formulae , P = (V*V)/R I get a huge number which is wrong because it is more than the power generated but when I use P = ( I * I ) / R , I get the correct answer.

Think I have a misconception about this [ P = (V*V)/R ] . Can someone tell me where I am wrong? and why so? Thanks in advance. =)
 
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The power equation is P = I*I*R (sub in ohm's law into the power equation)
 
Yah I think he typo'd there. What he's missing is that power generator is FORCING a certain power output, and thus FORCING a certain current through that potential difference, which you found correctly

However if that power generator wasn't limiting you, you would have 10kV across 5 ohms, which corresponds to a current flowing of 2000amps, which when put into P=IV, doing 2000*10kV gives you the incorrect huge answer you got

by using P=V^2/R you're finding the power dissipated in 5ohm wires with a potential difference of 10kV, ignoring the fact that you're limited to 100kW from the power generator. By finding the actual current through the wires, you incorporate that
 
Ohh.Thanks blochwave! Understood why I was wrong. =)
 
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