Using the simulation in class / Teaching why electrical power depends on current and voltage and how to measure it
Open in new tab:
HEADLINE LEARNING POINTS
Power is how quickly energy is shifted
Power is measured in watts (W)
The brighter the bulb, the higher the power
Electrical power (W) = current (A) x potential difference (V)
UNDERSTANDING
In our homes all the bulbs are connected across the same mains p.d. (apart from low voltage bulbs connected through a transformer)
We can use brightness as a rough proxy for how quickly energy is shifted in a bulb - but really it depends in a complex way on how the bulb is designed, as well as how we perceive brightness
With that proviso, bright bulbs in the home have a bigger current flowing (alternately back and forth) through them than dim ones
So bright bulbs have a lower resistance than dim ones - provided you’re connecting them across the same potential difference
This simple story is confused once we start changing the power supply voltage, as we tend to do at school
When we change the power supply voltage, we change the p.d. across the bulb AND the current through it
This changes how much energy each charge loses (potential difference) as it passes through the bulb, and how quickly charges with energy arrive at the bulb (current)
That’s why electrical power = current x voltage
In this case the bulb itself doesn’t have a property of being powerful - it is a function of the circuit it’s connected in
USING THE SIMULATION
Look at power qualitatively first by observing the current (speed of charges), the potential difference (amount of red stuff per black dot), and power (how quickly the red energy circle expands)
Change the battery voltage and ask these questions:
What happens to the speed of the charges?
What happens to how much energy they are carrying?
What happens to how quickly energy is spreading out at the bulb?
Does power depend on current only, potential difference only, or both?
Then connect the ammeter, voltmeter and power meter and show numerically how power = current x potential difference
Then keep the battery voltage constant and change the bulb resistance
Show that the potential difference across the bulb stays constant but the current changes
Ask:
If you keep the battery voltage constant, are high resistance bulbs bright or dim?
Do high resistance bulbs make the battery work harder or less hard?
Do high resistance bulbs run the battery down slower or faster?
SUBTLETIES
There’s no such thing as a power meter that you can attach in such a simple way as in our simulation
In the simulation the bulb is modelled as being Ohmic, but in reality filament bulbs significantly increase their resistance as the filament gets hotter
MISCONCEPTIONS
The most common misconception is that high-resistance bulbs are brighter than low-resistance ones - probably due to some combination of three reasons
1. People think that ‘the current’ has to struggle harder to make its way through a high resistance - the constant current misconception again
2. They know that kinked or damaged wires are a fire risk because of the increased resistance
But the explanation for these risks is not just to do with increasing resistance, but for much more subtle reasons involving resistors in series
Increasing the resistance of one of two resistances in series decreases the overall current but increases the p.d. across that resistance, because it takes a bigger share of the power supply voltage
If the resistance has a big proportional change but is still quite small overall, then the current stays quite high, but the p.d. increases a lot, so the power dissipated in that resistance increases significantly, even though the power supplied by the supply drops because it feels a bigger resistance overall
If the resistance gets too big then the current drops a lot and the increase in its share of the voltage isn’t enough to compensate, so less power is dissipated in it
This argument is way beyond most pre-16 courses, so we’re left with the incorrect notion that increasing resistance increases power dissipated
3. They recall that power = current squared x resistance, so assume that increasing resistance should increase power
But resistance and current are not independent - as resistance goes up, current goes down - and because current is squared, the effect of the decreasing current is greater than the increasing resistance