Exemplifying a calculation to give students good practice to emulate
What the activity is for
This is an interactive teacher demonstration with a difference. It involves working not with apparatus, but with numbers in carrying out circuit calculations. The idea here is to make explicit to students the steps, or underlying strategy, involved in making calculations. All too often this systematic guidance is missed out and students struggle to make sense of relatively simple questions.
What the activity is for
Here we suggest:
circuit 1: 12 volt supply and two resistors in series, 40 ohm and 80 ohm
circuit 2: 12 volt supply and two resistors in parallel, 48 ohm and 60 ohm
What happens during this activity
Set out below are the approaches to carrying out calculations for different kinds of circuit. These may be used as and when appropriate in a lesson sequence. It's a good idea to have apparatus to hand so that the steps in the calculation can be followed with the equipment.
Circuit 1: Two resistors in series
Question: What is the potential difference across the 80 ohm resistor?
In answering a question such as this, it is a good idea to talk through the overall approach with your class, making clear all of the key points:
With two resistors in series, the total voltage (12 volt) of the supply is shared between them.
Since the resistors are not of equal value, the voltage is shared with a greater potential difference across the bigger resistor. More energy is shifted as the current passes through the bigger resistor.
The current through both resistors is the same.
If we know the value of the electrical current then the potential difference across either resistor can be calculated using the equation: V = R × I.
Talking through the calculation
Teacher: So, how do we calculate the electric current in this circuit?
Sarah: From the voltage and resistance?
Teacher: Which resistance?
Will: The total resistance.
Teacher: Exactly! We can find the current in this loop from the voltage and total resistance using our old friend I = VR, so I = 12 volt40 ohm + 80 ohm, which simplifies to I = 12 volt120 ohm, for which you can work out that the current is 0.1 A.
Teacher: So the current in the loop is 0.1 ampere.
Teacher: We can now calculate the potential difference across the 80 ohm resistor: V = R × I, so the potential difference is 80 Ω × 0.1 A.
The potential difference across the 80 ohm resistor is equal to 8 volt.
Teacher: So what is the potential difference across the 40 ohm resistor?
Sam: Is it 4 volt?
Teacher: Why do you think that?
Sam: Because 8 from 12 volt leaves 4 volt.
Teacher: Yes! That's right. And why does that make good sense looking at the values of resistance?
The final point here is that there is twice the potential difference across the 80 ohm resistor compared with the 40 ohm resistor. In other words, the potential difference is shared in proportion to the resistance values.
Teacher: When you become expert at these calculations, like me(!), you'll be able just to glance at values such as these and give the answer straight away. Of course, the numbers aren't always as easy as this.
Discussing the calculations
Circuit 2: two resistors in parallel
Question: What is the current in the supply?
In answering a question such as this, it's a good idea to talk through the overall approach with your class, making clear all of the key points:
The current in each of the resistors can therefore be calculated using I = VR for each loop in turn, so I1 = VR1 and I2 = VR2.
Teacher: So, how do we calculate the current in the 48 ohm resistor?
John: From the voltage and resistance?
Teacher: And what is the voltage?
Will: 12 volt.
Teacher: Exactly! We can find the current in the 48 ohm resistor using the connection between I, V and R.
Using I1 = VR1, in this case 12 volt48 ohm, so the current works out at 0.25 A.
In the same way, the current through the 60 ohm resistor is calculated, Using I2 = VR2, in this case 12 volt60 ohm, so the current works out at 0.20 A.
The total current in the supply must therefore be the sum of these values: 0.25 ampere and 0.2 ampere. The current in the supply is 0.45 ampere.