When a circuit changes, follow the chain: resistance, then current, then potential difference across each part, then power. A good explanation names each quantity in order and gives a number or a direction for each.
This lesson belongs to integrated physical reasoning because it combines calculation with a written explanation, which is how circuit questions are usually asked.
How do I explain a circuit change step by step?
Always start with what the circuit did before the change, then what changed, then each consequence.
- Find the total resistance before and after. Series resistances add. Parallel resistances combine to a smaller total.
- Find the current from the cell using I = V ÷ R, with V the cell’s potential difference.
- Find the potential difference across each component using V = I × R, for the current through that component.
- Find the power using P = I × V for each component and for the whole circuit.
- Check conservation: the potential differences around a series loop add up to the cell’s value, and the branch currents add up to the total current.
Worked example (invented data)
A 6.0 V cell is connected in series with a 4.0 Ω resistor and a 2.0 Ω resistor. A second 4.0 Ω resistor is then added in parallel with the first 4.0 Ω resistor. Explain what happens to the current and to the power in the 2.0 Ω resistor.
Before the change. Total resistance = 4.0 + 2.0 = 6.0 Ω. Current I = 6.0 ÷ 6.0 = 1.0 A. Potential difference across the 2.0 Ω resistor = 1.0 × 2.0 = 2.0 V. Its power = 1.0 × 2.0 = 2.0 W.
After the change. The two 4.0 Ω resistors in parallel have total resistance 4.0 ÷ 2 = 2.0 Ω. Total resistance = 2.0 + 2.0 = 4.0 Ω.
Current from the cell = 6.0 ÷ 4.0 = 1.5 A.
Potential difference across the 2.0 Ω resistor = 1.5 × 2.0 = 3.0 V. Its power = 1.5 × 3.0 = 4.5 W.
Branch check. The parallel pair shares 6.0 − 3.0 = 3.0 V. Each 4.0 Ω branch carries 3.0 ÷ 4.0 = 0.75 A, and 0.75 + 0.75 = 1.5 A, which matches.
Power check. Total power = 6.0 × 1.5 = 9.0 W. The parts add to 4.5 + (0.75 × 3.0) + (0.75 × 3.0) = 4.5 + 2.25 + 2.25 = 9.0 W.
Written explanation. Adding the parallel resistor lowers the total resistance from 6.0 Ω to 4.0 Ω. The current from the cell rises from 1.0 A to 1.5 A. The same current now flows through the 2.0 Ω resistor, so its potential difference rises from 2.0 V to 3.0 V and its power rises from 2.0 W to 4.5 W.
The mistake to watch for
A student reasons that adding a component always makes the circuit “harder to push current through.”
Mistaken answer: “More resistors means more resistance, so the current falls and the 2.0 Ω resistor gets dimmer.”
That is true for a series addition. A parallel branch opens a second route, so total resistance falls and the current rises.
The correction is to ask first whether the new component is in series or in parallel with the old one. Draw it if you need to.
Check yourself
All data are invented.
1. A 12 V supply is connected across a 3.0 Ω resistor and a 6.0 Ω resistor in series. Find the current and the potential difference across the 6.0 Ω resistor.
Show answer
R = 9.0 Ω, so I = 12 ÷ 9.0 = 1.33 A, which is 1.3 A to 2 significant figures. Potential difference across 6.0 Ω = 1.333 × 6.0 = 8.0 V. The other resistor takes 4.0 V, and 8.0 + 4.0 = 12 V.
2. Two 6.0 Ω resistors are connected in parallel. What is their total resistance?
Show answer
Equal resistors in parallel give 6.0 ÷ 2 = 3.0 Ω. Check with 1/R = 1/6 + 1/6 = 2/6, so R = 3.0 Ω.
3. A lamp is rated 12 V, 24 W. Find its working current, its resistance, and the energy it transfers in 60 s.
Show answer
I = P ÷ V = 24 ÷ 12 = 2.0 A. R = V ÷ I = 12 ÷ 2.0 = 6.0 Ω. Energy = P × t = 24 × 60 = 1440 J.
Where this leads next
Move on to relating a nuclear or space model to supplied evidence, or return to combining energy and motion to see the same energy-transfer habit in a different setting. The scientific investigation critic helps when you design a circuit investigation and must justify what you control.
If your numbers are correct but your written explanations stay one step short, our teachers can work on that gap in online one-to-one Co-ordinated Sciences tuition.