The problems are on the ohm's law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Ohm's law says that the current through an ohmic conductor is proportional to the potential difference across it, so V = I × R. These worked answers correct the 20 problems of the grade 8 science worksheet on ohm's law, from quantities and units to a fire alarm sensor: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the final result in bold with its unit. Check that you convert milliamperes and kilohms before substituting.
Level: Grade 8 science (Year 9 · Class 8) · Curriculum: MS-PS2-3
Problem 1 Quantities and units
1. The unit is the ohm, symbol Ω.2. An ammeter, connected in series.
3. A voltmeter, connected in parallel across the component.
Problem 2 Prefix conversions
1. 4.7 × 1000 = 4700 Ω.2. 250 ÷ 1000 = 0.250 A.
3. 0.033 × 1000 = 33 mA.
4. 2200 ÷ 1000 = 2.2 kΩ.
Problem 3 Find the current
1. I = V ÷ R.2. I = 12 V ÷ 4.0 Ω = 3.0 A.
Problem 4 Find the voltage
1. V = I × R = 0.50 A × 24 Ω = 12 V.Problem 5 Find the resistance
1. R = V ÷ I = 9.0 V ÷ 0.30 A = 30 Ω.Problem 6 Where do the meters go?
1. The ammeter A is in series and the voltmeter V is in parallel with the resistor.2. V measures the potential difference across the resistor.
Problem 7 Two resistors on a graph
1. R = 8.0 ÷ 0.40 = 20 Ω.2. R = 8.0 ÷ 0.16 = 50 Ω.
3. X has the smaller resistance, so more current flows for each volt, and the line is steeper.
Problem 8 A kettle element
1. I = 230 ÷ 46 = 5.0 A.2. The current would double to 10 A, because I is inversely proportional to R.
Problem 9 A longer wire
1. R = 1.6 ÷ 0.80 = 2.0 Ω.2. Twice the length gives twice the resistance: 4.0 Ω.
3. I = 1.6 ÷ 4.0 = 0.40 A.
Problem 10 Data table
1. 2.0 ÷ 0.10 = 20, 4.0 ÷ 0.20 = 20, 6.0 ÷ 0.30 = 20, 8.0 ÷ 0.40 = 20: 20 Ω each time.2. R is constant, so the resistor is ohmic. Its I–V graph is a straight line through the origin.
Problem 11 A bulb’s resistance
1. R = 2.0 ÷ 0.30 = 6.7 Ω.2. R = 6.0 ÷ 0.50 = 12 Ω.
3. The filament gets hotter at the higher voltage, so its atoms vibrate more and the resistance rises.
Problem 12 A dimmer switch
1. I = 6.0 ÷ 5.0 = 1.2 A.2. I = 6.0 ÷ 20 = 0.30 A.
3. The current falls, so the lamp gets dimmer.
Problem 13 Spot the errors
1. (a) Doubling V doubles I for an ohmic resistor. (b) R stays the same. (c) A voltmeter is connected in parallel.Problem 14 Electrical safety
1. I = 230 ÷ 100 000 = 0.0023 A = 2.3 mA.2. I = 230 ÷ 1 000 = 0.23 A = 230 mA.
3. The current is 100 times bigger with wet skin, which can be fatal because the heart is affected by currents above about 30 mA.
Problem 15 Wire comparison
1. Q is twice as long, so R is 8.0 Ω.2. S has double the area, so R is halved: 2.0 Ω.
Problem 16 An LED resistor
1. The voltages in series add up to 9.0 V, so 9.0 − 2.0 = 7.0 V.2. 20 mA = 0.020 A.
3. R = 7.0 ÷ 0.020 = 350 Ω.
Problem 17 Two resistors in series
1. R = 100 + 220 = 320 Ω.2. I = 9.6 ÷ 320 = 0.030 A.
3. V = 0.030 × 220 = 6.6 V.
4. Across 100 Ω: 0.030 × 100 = 3.0 V, and 3.0 + 6.6 = 9.6 V, which matches.
Problem 18 Gradient of a graph
1. Gradient = 0.20 A ÷ 8.0 V = 0.025 A/V.2. R = 1 ÷ gradient = 1 ÷ 0.025 = 40 Ω.
3. I = 9.0 ÷ 40 = 0.225 A, about 0.23 A.
Problem 19 Plan an investigation
1. Independent: length. Dependent: resistance. Controls: same wire, same thickness, same temperature (keep the current small).2. Connect the wire in series with an ammeter and a cell, put a voltmeter across the length, read V and I, and calculate R = V ÷ I for lengths from 10 cm to 100 cm.
3. A straight line through the origin: R is proportional to length, because doubling the length doubles the number of collisions.
Problem 20 A fire alarm sensor
1. At 20 °C: R = 5.0 ÷ 0.0010 = 5000 Ω. At 60 °C: R = 5.0 ÷ 0.0040 = 1250 Ω.2. 5000 ÷ 1250 = 4.0, so the resistance falls to a quarter.
3. When heat raises the temperature the current rises sharply, which a circuit can detect to trigger the alarm.
