Electricity Tutorial 5 - Energy and Power in Circuits
Contents |
Conventional Current
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:
Power in a Current
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 P against I^{2} we would get a straight line graph.
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.
Buying Electrical Energy (OCR)
In the early days of nuclear power, some people said that electricity would be in such plentiful supply that it would not be necessary to meter it. Back to reality. It costs a lot of money to generate and supply energy, and it need to be paid for.
Electricity is sold by the kilowatt-hour (kW h) which is defined as:
The amount of energy used by an appliance taking a power of 1 kW running continuously for 1 hour.
Not kilowatt per hour (kW/h) |
Convert 1 kW h to the equivalent energy in joules (J) |
The kilowatt-hour is often referred to as a unit on an electricity bill. The average cost per unit as this is being written is about 14 p.
On all appliances there is a plate that shows the power consumption. This one is on a tumble dryer.
This tumble dryer takes a power of 2300 W.
It takes 75 minutes to dry a load of washing. How much does this cost? |
In the electricity bill, there is also a standing charge. This covers the cost of the infrastructure to supply your house with electricity. It is the same regardless of how much or how little electricity you use. Depending how you pay for your electricity, the standing charge is about 22 p a day. This works out at about £20 a quarter (90 days).
At the start of a quarterly billing period (3 months or 90 days) the electricity meter reads 025628 kW h. At the end it reads 026344 kW h. If electricity costs 14 p per unit, calculate the electricity including the standing charge (£22 a quarter for this supplier). |
Large users of electrical energy pay in units of megawatt-hours (MW h) and the cost is about £90 per megawatt hour (9 p per kilowatt-hour).