__
Electricity Tutorial 5 -
Energy
and Power in Circuits__

**Conventional** current goes from **positive**
to **negative**. **Electrons** carry energy around the circuit; they
go from **negative** to **positive**. In the early days, physicists
didn't know about the electron, which is why they got it all wrong.
Correction would require a complex re-write of the Laws of Physics, a task which
no-one is likely to want to tackle. So all conventional currents
are from positive to negative. All currents are treated as **conventional**.

We can measure the energy in a circuit by measuring the voltage and the current.

The voltage current graph looks like this:

Suppose
a current
*I*
amps flows for
*t *
seconds in a component. The charge that
flowed led to *E* joules being dissipated in the component.

We know that:

So if we substitute *Q*
out of the second equation, we get:

Now

Power = __energy__

time

So we can write:

It
doesn't take a genius to see that the term
*t* cancels out to leave us
with:

*P
= IV *

Power
is measured in **watts** (W). 1 watt = 1 joule per second

** **

A 12 volt heater takes a current of 3.6 A. It is left to heat up an aluminium block for a period of 45 minutes. How much heat energy is transferred to the aluminium block? |
||

What current is consumed by a 60 W light bulb operating on the 230 V mains? |

__The
Heating Effect of a Current__

We know that:

So we can write:

*
P = I × IR*

So it doesn’t take a genius to see that by substituting the second equation into the first, we get:

We know that:

So we can write:

*
P = V × **V/R*

So it doesn’t take a genius to see that by substituting the second equation into the first, we get:

The graph looks like this:

The graph shows that if we
double the current, we get four times the power, consistent with the idea that
*P*
µ* I ^{2}*.
If we were to plot

* *

A resistor of value 50 ohms is rated at 1 watt. This means that if it has to give out more power than 1 watt, it will start to get hot. What is the maximum current that it can handle? |
||

The same resistor as in Q 9 is then connected to a 20 volt supply. What power will it dissipate now? What do you think will happen to the resistor? |

The picture shows the heating effect of a current on a resistor!

This was a 33 W resistor connected to a 20 V supply. The current would be 20 V ÷ 33 W = 0.61 A

The power would be 0.61 A × 20 V = 12 watts. Plenty enough to fry a 1 watt resistor.

It is important that we ensure that any current limiting resistors can dissipate the power through them. The above situation could be highly dangerous.