The problems are on the ohm's law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Current is the flow of charge in amperes, potential difference is the energy per coulomb in volts, and resistance in ohms is the opposition to current. These worked answers correct the 20 problems of the Physics 9-10 worksheet on ohm's law, from quantities and units to mystery components: each solution shows the formula, the substitution and the final answer with its unit.
Full answers follow each problem in the same order. Check each substitution, confirm the units cancel, and compare any graph readings with the values given in the figures.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE P2, IGCSE 4.3
Problem 2 Finding voltage
1. V = I × R.2. V = 0.50 × 12 = 6.0 V.
Problem 3 Finding resistance
1. R = V ÷ I.2. R = 9.0 ÷ 0.30 = 30 Ω.
Problem 4 Finding current
1. I = V ÷ R.2. I = 12 ÷ 48 = 0.25 A.
Problem 5 Unit conversions
1. 350 mA = 0.35 A.2. 4.7 kΩ = 4700 Ω.
3. 2.2 MΩ = 2 200 000 Ω (2.2 × 10⁶ Ω).
Problem 6 Reading a resistor graph
1. Halfway between 0.20 A and 0.40 A: 0.30 A.2. R = 3.0 ÷ 0.30 = 10 Ω (the same at any point).
3. The resistance is constant and I is proportional to V, so the resistor is ohmic.
Problem 7 Thermistor resistance
1. 3.0 mA = 0.0030 A.2. R = 6.0 ÷ 0.0030 = 2000 Ω = 2.0 kΩ.
Problem 8 Energy from the battery
1. W = Q × V.2. W = 45 × 12 = 540 J.
Problem 9 Charge and energy in a lamp
1. Q = I t = 0.50 × 120 = 60 C.2. W = Q V = 60 × 6.0 = 360 J.
Problem 10 Filament lamp data
1. R = V ÷ I: 10.0 Ω, 11.1 Ω, 12.5 Ω, 13.8 Ω, 15.2 Ω.2. The resistance increases as the current and voltage rise.
3. The filament gets hotter, so its ions vibrate more and collide with electrons more often.
Problem 11 A diode in a phone charger
1. About 0.4 V (the current rises steeply above about 0.6 V).2. R = 0.75 ÷ 0.050 = 15 Ω.
3. Almost zero: the resistance is very large in reverse.
Problem 12 LDR street lamp
1. I = 4.0 ÷ 200 = 0.020 A (20 mA).2. I = 4.0 ÷ 50 000 = 8.0 × 10⁻⁵ A (80 µA).
3. 0.020 ÷ 0.000080 = 250 times.
Problem 13 Thermistor graph
1. 5.3 kΩ at 40 °C and 1.3 kΩ at 80 °C.2. At 40 °C: I = 6.0 ÷ 5300 = 1.1 mA. At 80 °C: I = 6.0 ÷ 1300 = 4.6 mA.
3. The resistance falls as temperature rises, so the current increases.
Problem 14 Resistance of wires
1. Three times as long gives three times the resistance: 12 Ω.2. Double the area halves it: 12 ÷ 2 = 6.0 Ω.
Problem 15 Spot the errors
1. An ammeter is connected in series.2. R = V ÷ I.
3. An NTC thermistor’s resistance decreases when it is heated.
4. A diode conducts in one direction only.
Problem 16 Gradient of an unknown resistor
1. Gradient = (101 − 21) ÷ (5.0 − 1.0) = 80 ÷ 4.0 = 20 mA/V = 0.020 A/V.2. R = 1 ÷ gradient = 1 ÷ 0.020 = 50 Ω.
3. Yes: the points lie on a straight line through the origin (current ∝ voltage), within experimental error.
Problem 17 Thermistor and fixed resistor
1. R = 5.3 + 1.0 = 6.3 kΩ; I = 6.0 ÷ 6300 = 0.95 mA.2. R = 1.3 + 1.0 = 2.3 kΩ; I = 6.0 ÷ 2300 = 2.6 mA.
3. V = I R: at 40 °C 0.95 mA × 1000 = 0.95 V; at 80 °C 2.6 mA × 1000 = 2.6 V. The voltage across the resistor rises with temperature, so an alarm can trigger when it passes a set value.
Problem 18 Hot and cold lamp
1. R = 12 ÷ 0.50 = 24 Ω.2. I = 12 ÷ 3.0 = 4.0 A, eight times the running current.
3. The large surge heats and stresses the thin filament, so a weak spot can melt at switch-on.
Problem 19 Plan a light investigation
1. Independent: distance from the lamp. Dependent: resistance. Control: lamp brightness and room light.2. Connect the LDR in series with an ammeter and a cell, with a voltmeter across the LDR. At each distance, calculate R = V ÷ I.
3. Draw a graph of R against distance. Keep hands away from the hot lamp.
Problem 20 Mystery components
1. P: 20 Ω, 20 Ω, 20 Ω. Q: 10 Ω, 13 Ω, 17 Ω. R: 40 Ω, 27 Ω, 15 Ω.2. P is the ohmic resistor (constant R), Q is the filament lamp (R rises as it heats) and R is the thermistor (R falls as it warms).
