These eight original questions follow electrical reasoning from diagrams and measurements. Assume ideal supplies. Attempt each one, including the written reason, before you open the answer.
Questions
1. A 6 V supply is connected to a 4 Ω resistor and an 8 Ω resistor in series. Find the potential difference across each.
Answer
Total resistance = 12 Ω, so I = 6 ÷ 12 = 0.50 A. Across 4 Ω: 0.50 × 4 = 2 V. Across 8 Ω: 0.50 × 8 = 4 V. Check: 2 + 4 = 6 V.
2. A 12 V supply is connected to a 6 Ω resistor and a 12 Ω resistor in parallel. Find the current in each branch and the total current.
Answer
Each branch has 12 V. The 6 Ω branch: 12 ÷ 6 = 2 A. The 12 Ω branch: 12 ÷ 12 = 1 A. Total = 3 A.
3. In question 2, the 12 Ω branch opens. State the new total current and the current in the 6 Ω branch.
Answer
The 6 Ω branch still has 12 V across it, so its current stays 2 A. The total current falls to 2 A, because only one branch is left.
4. A filament lamp gives 1.5 V with 0.30 A, and 6.0 V with 0.60 A. Calculate V ÷ I each time and explain the change.
Answer
At 1.5 V: 1.5 ÷ 0.30 = 5 Ω. At 6.0 V: 6.0 ÷ 0.60 = 10 Ω.
The larger current heats the filament, and a hotter filament has a higher resistance. So the lamp is not ohmic.
5. An 800 W heater runs for 15 minutes. Find the energy in joules and in kWh.
Answer
Time = 15 × 60 = 900 s, so E = 800 × 900 = 720 000 J.
In kWh: 0.8 kW × 0.25 h = 0.20 kWh.
6. Which uses more energy: a 2 kW iron used for 10 minutes, or a 0.5 kW fan used for 1 hour?
Answer
Iron: 2 × (10 ÷ 60) = 0.333 kWh. Fan: 0.5 × 1 = 0.50 kWh.
The fan uses more energy, even though the iron has the higher power, because it runs six times as long.
7. A 6 V supply is connected in series to a 10 Ω lamp and a rheostat. With the rheostat at 2 Ω, then at 5 Ω, find the potential difference across the lamp each time.
Answer
At 2 Ω: I = 6 ÷ 12 = 0.50 A, so V = 0.50 × 10 = 5.0 V. At 5 Ω: I = 6 ÷ 15 = 0.40 A, so V = 0.40 × 10 = 4.0 V.
A larger rheostat resistance lowers the current, so the lamp’s potential difference falls.
8. A student says that in a series circuit the current is used up by the lamp, so an ammeter after the lamp reads less. Explain the error.
Answer
In a series circuit there is only one path, so the same current passes through every point. The lamp transfers energy, but it does not use up charge.
The ammeter after the lamp reads the same value as the ammeter before it.
If you got these wrong
For questions 1 and 7, revisit comparing potential difference readings. For questions 2 and 3, see predicting what an open branch changes. For question 4, see explaining a non-linear current-voltage graph.
For questions 5 and 6, see separating electrical energy from power. Log each slip in the mistake log and paper-error review.
For a teacher to go through your answers, see online one-to-one Physics tuition or try the one-hour trial class (from RM50).