LessonPower and energy · 20 min
Practice set: Ohm's law, power and energy
Work these with a pencil and a calculator. Write the formula first, then the numbers with units, then the answer - and ask whether its size makes sense. The challenges take several steps, like real design work.
- V, I, R, P
- volts, amps, ohms, watts
- E
- energy: J (t in s) or Wh (t in h)
Practice 1What current flows through 330 Ω with 5 V across it?
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Answer: 15.15 mA
I = 5 / 0.33 kΩ = 15.2 mA.
Practice 2A 12 V car bulb takes 4 A. What is its power?
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Answer: 48 W
P = 12 × 4 = 48 W.
Practice 3What is the resistance of a 230 V 1500 W heater?
Hint
R = V² / P.
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Answer: 35.27 Ω
R = 230² / 1500 = 52 900 / 1500 = 35.3 Ω.
Practice 4A resistor gets 0.5 W with 10 V across it. What is its resistance?
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Answer: 200 Ω
R = V² / P = 100 / 0.5 = 200 Ω.
Practice 5A router takes 12 V at 1 A all day and all night. How many kWh in 30 days?
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Answer: 8.64 kWh
12 W × 24 h × 30 = 8640 Wh = 8.64 kWh.
Practice 6Which uses more energy: a 1000 W iron for 15 minutes or a 60 W fan for 5 hours?
- The iron
- The fan
- Both the same
- It cannot be worked out
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Answer: B. The fan
Iron: 1000 × 0.25 = 250 Wh. Fan: 60 × 5 = 300 Wh. The fan uses more.
Challenge 7A 5 V USB supply feeds a 3 m cable whose two wires have 0.15 Ω each. A device takes 2 A. What voltage reaches the device?
Hint
The current flows through both wires: 0.3 Ω in all.
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Answer: 4.4 V
Drop = 2 × 0.3 = 0.6 V, so 5 − 0.6 = 4.4 V reach the device - why thin, cheap USB cables charge slowly.
Challenge 8A 12 V 18 W lamp and a 12 V 6 W lamp are connected in series across 12 V. Assuming fixed resistances, what power does the 6 W lamp take?
Hint
Find each lamp's resistance (R = V² / P), then the current in series, then I² R.
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Answer: 3.375 W
R18 = 144 / 18 = 8 Ω, R6 = 144 / 6 = 24 Ω. I = 12 / 32 = 0.375 A. P6 = 0.375² × 24 = 3.375 W - more than the "bigger" lamp's 1.125 W. In series, the higher resistance takes more power.
Challenge 9A sensor node sleeps at 20 µA for 59 s, then wakes at 80 mA for 1 s, every minute. Running from a 2400 mAh battery, about how many days does it last (ideal)?
Hint
Average current = (20 µA × 59 + 80 mA × 1) / 60.
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Answer: 74 days
Average = (0.02 mA × 59 + 80 mA × 1) / 60 = (1.18 + 80) / 60 = 1.353 mA. 2400 / 1.353 = 1774 h = 73.9 days. Cut the awake time, not the sleep current, to gain most.