# Power

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## Power and Energy Transfer

Energy transfers occur in electrical appliances. The amount of energy transferred depends on the power of the appliance and how long it is switched on for. Different appliances have different power ratings, which are the maximum power the appliance may be operated at.

## Transferring Energy

Appliances transfer electrical energy to useful energy. For example, a fan transfers electrical energy to kinetic energy of the fan and an oven converts electrical energy to thermal energy to heat up your food.

The amount of energy transferred in an appliance can be calculated using the following equation:

$\textcolor{aa57ff}{E=Pt}$

• $E=$ energy in joules $\text{(J)}$
• $P=$ power in watts $\text{(W)}$
• $t=$ time in seconds $\text{(s)}$

This equation tells us that more energy is transferred if the appliance has a higher power rating or if it is switched on for a longer time.

Power rating is the maximum power an appliance can safely operate at. This means the power rating of an appliance tells us the maximum energy transferred per second by an appliance.  However, having a higher power rating does not mean that more useful energy is transferred. A high power rating appliance may not be as efficient as a low power rated appliance.

Energy transferred can also be calculated by working out how much work has been done by the charge carriers that have flowed through the circuit across a potential difference:

$\textcolor{aa57ff}{E=QV}$

• $E=$ energy in joules $\text{(J)}$
• $Q=$ charge in coulombs $\text{(C)}$
• $V=$ potential difference in volts $\text{(V)}$.
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## Calculating Power

Recall that power can be calculated using the equation:

$P=VI$

• $P=$ power in watts $\text{(W)}$
• $V=$ potential difference in volts $\text{(V)}$
• $I=$ current in amps $\text{(A)}$.

If you don’t know the potential difference, you can also calculate power using:

$P=I^2R$

• $P=$ power in watts $\text{(W)}$
• $I=$ current in amps $\text{(A)}$
• $R=$ resistance in ohms $(\Omega)$
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## Example: Calculating Charge Flow

A lightbulb of power $15\text{ W}$ is switched on for $2\text{ minutes}$. If the potential difference across the lightbulb is $10\text{ V}$, how much charge has flowed through the bulb?

[3 marks]

$\bold{\text{E}=\text{Pt}}=\textcolor{d11149}{15\text{ W}}\times \textcolor{f95d27}{(2\times60)\text{ s}}=1800\text{ J} \\ \bold{\text{E}=\text{QV}}$

Rearrange for Q:

$\text{Q}=\dfrac{\text{E}}{\text{V}}=\dfrac{1800\text{ J}}{\textcolor{7cb447}{10\text{ V}}}=\bold{180\text{ C}}$

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## Power Example Questions

Electrical energy is converted to heat energy in the heater, this transfers to heat energy in the air.

Gold Standard Education

$t = 60 \times 60 \text{ s} \\ \bold{\text{E}=\text{Pt}}=5\times(60 \times 60)\text{ s}=\bold{18\,000\text{ J}}$

Gold Standard Education

$\bold{\text{E}=\text{QV}}$

So $\text{Q}=\dfrac{\text{E}}{\text{V}}=\dfrac{168\text{ J}}{12\text{ V}}=\bold{14\text{ C}}$

Gold Standard Education

$\bold{\text{P}=\text{I}^2\text{R}}=(2\text{ A})^2\times 10 \Omega=\bold{40\text{ W}}$

Gold Standard Education

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