# Power Output (P=VI) Watch

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Power is supplied from a power station at a potential difference of 120 kV. The power output of the station is 60 MW and is fed through cables of resistance 4.0 Ω to a town. Calculate potential difference supplied to the town.

So the solution given is

I(from power station) = P/V = 60MW / 120kV = 500A

V(at town) = P/I = (59 x 106) / (500A) = 118 kV

What I don't understand is why the current at the town remains the same as from the power station (500A), seeing as the current travels through cables with resistance. Could anyone explain this to me?

Also, I'm confused as to whether P=IV is the electric power or power dissipated?

So the solution given is

I(from power station) = P/V = 60MW / 120kV = 500A

V(at town) = P/I = (59 x 106) / (500A) = 118 kV

What I don't understand is why the current at the town remains the same as from the power station (500A), seeing as the current travels through cables with resistance. Could anyone explain this to me?

Also, I'm confused as to whether P=IV is the electric power or power dissipated?

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#2

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What I don't understand is why the current at the town remains the same as from the power station (500A), seeing as the current travels through cables with resistance. Could anyone explain this to me?

**RyanCWY**)What I don't understand is why the current at the town remains the same as from the power station (500A), seeing as the current travels through cables with resistance. Could anyone explain this to me?

Also, I'm confused as to whether P=IV is the electric power or power dissipated?

The power supplied by the generator is being dissipated by the load - the homes that it supplies and that lost in the cables. In this example, we lose 2kV*500A=1MW in the cables, with 59MW being used by the supplied homes.

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(Original post by

The current doesn't change as it travels through a resistance - just like water in your home's plumbing - what goes in must come out.

What's the difference? You'll get the power dissipated for the bit corresponding to the voltage drop that you use.

The power supplied by the generator is being dissipated by the load - the homes that it supplies and that lost in the cables. In this example, we lose 2kV*500A=1MW in the cables, with 59MW being used by the supplied homes.

**RogerOxon**)The current doesn't change as it travels through a resistance - just like water in your home's plumbing - what goes in must come out.

What's the difference? You'll get the power dissipated for the bit corresponding to the voltage drop that you use.

The power supplied by the generator is being dissipated by the load - the homes that it supplies and that lost in the cables. In this example, we lose 2kV*500A=1MW in the cables, with 59MW being used by the supplied homes.

And regarding the last part about dissipation, I think I don't think i fully understand what you mean. For example, if the question was

A battery of e.m.f. 6V and internal resistance of 0.4ohm is connected to an external resistor of 2.9ohms. What is the power supplied to the external resistor?

How do I solve this? I think that I have trouble in understanding when the P in P=VI means power loss to heat and when it means power supplied

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So does that mean that for a current travelling through a wire with fixed resistance (no temperature change etc), in the formula V=IR, I and R are constants?

**RyanCWY**)So does that mean that for a current travelling through a wire with fixed resistance (no temperature change etc), in the formula V=IR, I and R are constants?

And regarding the last part about dissipation, I think I don't think i fully understand what you mean. For example, if the question was

A battery of e.m.f. 6V and internal resistance of 0.4ohm is connected to an external resistor of 2.9ohms. What is the power supplied to the external resistor?

A battery of e.m.f. 6V and internal resistance of 0.4ohm is connected to an external resistor of 2.9ohms. What is the power supplied to the external resistor?

The total resistance is 3.3 Ohms, from which you can calculate the current that flows (6/3.3 A). I'd then use to calculate the power converted by the external resistor. You could also calculate the voltage drop across the external resistor and then use , where V is the voltage across the external resistor, i.e. .

How do I solve this? I think that I have trouble in understanding when the P in P=VI means power loss to heat and when it means power supplied

In your power station question, the power dissipated by the supply lines isn't useful for the homes, so is said to be lost. If the power dissipated in the home is put through a resistance to generate heat, we class it as useful, but it's still just electricity being converted to heat - it's just a matter of where.

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#5

(Original post by

Power is supplied from a power station at a potential difference of 120 kV. The power output of the station is 60 MW and is fed through cables of resistance 4.0 Ω to a town. Calculate potential difference supplied to the town.

So the solution given is

I(from power station) = P/V = 60MW / 120kV = 500A

V(at town) = P/I = (59 x 106) / (500A) = 118 kV

What I don't understand is why the current at the town remains the same as from the power station (500A), seeing as the current travels through cables with resistance. Could anyone explain this to me?

Also, I'm confused as to whether P=IV is the electric power or power dissipated?

**RyanCWY**)Power is supplied from a power station at a potential difference of 120 kV. The power output of the station is 60 MW and is fed through cables of resistance 4.0 Ω to a town. Calculate potential difference supplied to the town.

So the solution given is

I(from power station) = P/V = 60MW / 120kV = 500A

V(at town) = P/I = (59 x 106) / (500A) = 118 kV

What I don't understand is why the current at the town remains the same as from the power station (500A), seeing as the current travels through cables with resistance. Could anyone explain this to me?

Also, I'm confused as to whether P=IV is the electric power or power dissipated?

I believe this kind of circuit is called a voltage divider. Type it into google if you want to find similar problems like the one you got.

Find the current which would be 500 ohms, by using P = IV

All you need to do is to find the voltage drop from the cables by using V = IR.

You will get a result of 2000v. Then to find the voltage drop from the town just subtract 120000v from 2000v and you will get 118000v.

Current doesn't drop as the voltage drops over the resistors instead (as beautifuly explained by RogerOxon).

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(Original post by

I'm not sure that I understand you. Think of a wire as a pipe - the current is the water flow rate and the voltage (drop) is the pressure drop across the pipe. You can have more flow (current), but it'll require a larger pressure drop to push it through.

The internal and external resistors both dissipate power, in the form of heat. It's just that we define that taken by the external resistor as useful.

The total resistance is 3.3 Ohms, from which you can calculate the current that flows (6/3.3 A). I'd then use to calculate the power converted by the external resistor. You could also calculate the voltage drop across the external resistor and then use , where V is the voltage across the external resistor, i.e. .

It's the power loss when the V is across a component that isn't 'useful'. It's the power supplied when the power does something that we want.

In your power station question, the power dissipated by the supply lines isn't useful for the homes, so is said to be lost. If the power dissipated in the home is put through a resistance to generate heat, we class it as useful, but it's still just electricity being converted to heat - it's just a matter of where.

**RogerOxon**)I'm not sure that I understand you. Think of a wire as a pipe - the current is the water flow rate and the voltage (drop) is the pressure drop across the pipe. You can have more flow (current), but it'll require a larger pressure drop to push it through.

The internal and external resistors both dissipate power, in the form of heat. It's just that we define that taken by the external resistor as useful.

The total resistance is 3.3 Ohms, from which you can calculate the current that flows (6/3.3 A). I'd then use to calculate the power converted by the external resistor. You could also calculate the voltage drop across the external resistor and then use , where V is the voltage across the external resistor, i.e. .

It's the power loss when the V is across a component that isn't 'useful'. It's the power supplied when the power does something that we want.

In your power station question, the power dissipated by the supply lines isn't useful for the homes, so is said to be lost. If the power dissipated in the home is put through a resistance to generate heat, we class it as useful, but it's still just electricity being converted to heat - it's just a matter of where.

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