# Power Dissipated: Find Solution & Understand Why

• AirForceOne
In summary, power lost in a 3 ohm wire resistance is I^2(R), where R is the resistance of the wire. If the voltage is increased, the current flow is decreased, with associated less wire losses.
AirForceOne

## Homework Statement

A power station delivers 750 kW of power at 12,000 V to a factory through wires with total resistance 3.0 ohms. How much less power is wasted if the electricity is delivered at 50,000 V rather than 20,000 V?

## Homework Equations

P = IV = (I^2)(R) = (V^2)/R

## The Attempt at a Solution

The answer is P_dissipated = (I^2)(R) = ((P^2)(R)/(V^2), but can anyone tell me why I can't use P = (V^2)/R? I think there's a gap in my understanding.

The power lost is I^2(R), where R is the resistance of the wire, and I is determined from I = P/V. If using P =V^2/'R', the 'R' in this formula is not the resistance of the wire, it is the 'equivalent' resistance of the factory equipment load.

PhanthomJay said:
The power lost is I^2(R), where R is the resistance of the wire, and I is determined from I = P/V. If using P =V^2/'R', the 'R' in this formula is not the resistance of the wire, it is the 'equivalent' resistance of the factory equipment load.

That's what I sort of thought. Would the equivalent resistance be the resistance of the wire PLUS the resistance of the factory equipment load?

AirForceOne said:
That's what I sort of thought. Would the equivalent resistance be the resistance of the wire PLUS the resistance of the factory equipment load?
That is sort of correct, except it's somewhat of an over simplification to call the factory load a resistance, since it's more of an 'impedance' , but nonetheless, the station must generate enough power to overcome the wire losses plus deliver enough load to the factory. I guess it's better to look at this way...if the factory needs 750 kW of power at its end (before step down transformers to usable low voltages, with associated additional losses) to run its equipment, lights, etc., then since power losses over the wires are I^2(R) or about 12 kW, the power station must supply 762 kW at its end in order for 750 kW to end up he factory end. If the delivery voltage is increased, the current flow is decreased, with associated less wire losses.

PhanthomJay said:
That is sort of correct, except it's somewhat of an over simplification to call the factory load a resistance, since it's more of an 'impedance' , but nonetheless, the station must generate enough power to overcome the wire losses plus deliver enough load to the factory. I guess it's better to look at this way...if the factory needs 750 kW of power at its end (before step down transformers to usable low voltages, with associated additional losses) to run its equipment, lights, etc., then since power losses over the wires are I^2(R) or about 12 kW, the power station must supply 762 kW at its end in order for 750 kW to end up he factory end. If the delivery voltage is increased, the current flow is decreased, with associated less wire losses.

I understand the part about power being lost by the wire but I am still scratching my head.

Power is just the rate of energy transfer from electrical energy to whatever other form of energy, but how can I picture that in a circuit diagram?

Would the circuit diagram look something like this?

Sorry, I am really confused!

Last edited:
Well of course this diagram is a very basic simplification, but it's more or less correct with the following corrections:

First, The problem wording is a bit unclear, because I'm not sure if 750 kW is being sourced from the power station, and 738 kW, more or less, ends up at the factory, due to the line losses in the 3 ohm wire resistance, or whether 750 kW ends up at the factory, in which case 762 kW of power, more or less, needs to be sourced from the power station, due to those line losses, so that 750 kW ends up at the factory end. I've been assuming this latter scenario.

Second, the voltage source of 12,000 volts exists at the station itself, not before it.

Third, the ground return is usually through earth, not hard-wired back to the source.

I'm not an electrical expert, though. This is about the extent of my electrical knowledge.

## 1. What is power dissipation?

Power dissipation refers to the amount of electrical energy that is converted into heat energy due to the resistance of a component or material. This heat can cause a decrease in the performance or lifespan of the component.

## 2. How is power dissipation calculated?

Power dissipation is calculated by multiplying the voltage drop across a component by the current flowing through it. This is known as Joule's Law: P = I^2R or P = VI, where P is power in watts, I is current in amperes, and R is resistance in ohms.

## 3. What factors affect power dissipation?

The main factors that affect power dissipation are the voltage applied, the current flowing through the component, and the resistance of the material. Other factors such as ambient temperature and heat dissipation capabilities can also play a role.

## 4. How is power dissipation related to efficiency?

In most cases, power dissipation decreases the efficiency of a system or device. This is because some of the electrical energy is being lost as heat instead of being used for its intended purpose. However, there are some cases where power dissipation is intentionally used, such as in heating elements or resistors.

## 5. How can power dissipation be reduced?

Power dissipation can be reduced by using materials with lower resistance, increasing the efficiency of the system, or implementing cooling methods such as heatsinks or fans. It is important for engineers and scientists to consider power dissipation in their designs to optimize performance and prevent overheating.

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