The problems are on the electrical power worksheet. Try them on your own first, then compare with the answers below.
Quick answer: In a series circuit the current is the same everywhere and the voltages add up to the supply; in a parallel circuit every branch has the full supply voltage and the branch currents add up. These worked answers correct the 20 problems of the Physics 9-10 worksheet on electrical power, from series or parallel? to design a lamp circuit: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. Final results are in bold. If your numbers differ slightly, check your rounding: intermediate values are kept unrounded and only the final result is rounded.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE P2, HS-PS3-3
Problem 1 Series or parallel?
1. Series (one loop, one current).2. Parallel (each branch is across the supply).
3. Parallel: the other branches still form complete loops.
4. Series: the resistances add, so the total is larger than any one. In parallel it would be smaller.
Problem 2 Phone charger power
1. P = I V = 5.0 × 2.0 = 10 W.2. t = 30 × 60 = 1800 s. E = P t = 10 × 1800 = 18 000 J.
3. E = 10 W × 0.5 h = 5.0 Wh (18 000 J ÷ 3600 = 5.0 Wh).
Problem 3 Three resistors in series
1. R = 15 + 25 + 10 = 50 Ω.2. I = V ÷ R = 9.0 ÷ 50 = 0.18 A.
3. V = I R = 0.18 × 25 = 4.5 V.
Problem 4 Reading labels
1. 2.5 × 1000 = 2500 W.2. 450 ÷ 1000 = 0.45 W.
3. 90 ÷ 60 = 1.5 h.
4. 1 kWh = 3.6 MJ, so 3.6 MJ = 1.0 kWh.
Problem 5 Voltage in a series loop
1. The voltages add to 6.0 V: V₂ = 6.0 − 2.2 = 3.8 V.2. I = V ÷ R = 2.2 ÷ 11 = 0.20 A.
3. The same 0.20 A flows through B: R = 3.8 ÷ 0.20 = 19 Ω.
4. P = V I = 3.8 × 0.20 = 0.76 W. (Check: 11 + 19 = 30 Ω and 6.0 ÷ 30 = 0.20 A.)
Problem 6 Currents in branches
1. A₁ measures the total: 0.80 + 1.50 = 2.30 A.2. Both lamps are across the cell: 9.0 V.
3. R₂ = 9.0 ÷ 0.80 = 11.25, so 11 Ω; R₃ = 9.0 ÷ 1.50 = 6.0 Ω.
4. P = V I = 9.0 × 2.30 = 20.7 W = 21 W (2 s.f.).
Problem 7 Toaster fuse
1. I = P ÷ V = 800 ÷ 240 = 3.3 A.2. A 3 A fuse would blow in normal use, and 13 A is too big to protect the flex, so choose 5 A: just above the working current.
3. t = 4 × 60 = 240 s. E = P t = 800 × 240 = 192 000 J.
Problem 8 Night fan costs
1. E = 0.060 kW × 8 h = 0.48 kWh.2. 0.48 × 30 = 14.4 kWh.
3. Cost = 14.4 × 1.50 = 21.60 cedis.
Problem 9 Two resistors in parallel
1. 1/R = 1/12 + 1/4 = 1/12 + 3/12 = 4/12 = 1/3, so R = 3.0 Ω.2. I = 9.0 ÷ 3.0 = 3.0 A.
3. I₁ = 9.0 ÷ 12 = 0.75 A; I₂ = 9.0 ÷ 4 = 2.25 A. They add up to 3.0 A.
Problem 10 Heating coil
1. P = I²R = 5.0² × 46 = 25 × 46 = 1150 W.2. V = I R = 5.0 × 46 = 230 V.
3. t = 120 s. E = P t = 1150 × 120 = 138 000 J.
Problem 11 Resistance of an iron
1. R = V² ÷ P = 220² ÷ 1000 = 48 400 ÷ 1000 = 48.4 Ω.2. I = P ÷ V = 1000 ÷ 220 = 4.5 A.
3. The normal current 4.5 A is larger than 3 A, so a 3 A fuse would melt every time the iron is switched on; a 5 A or 13 A fuse is needed.
Problem 12 Why parallel at home?
1. Each lamp gets the full mains voltage so it shines at its rated brightness, and each lamp can be switched on or off separately.2. In series the circuit would be broken, so all the lamps would go out.
3. In parallel the lamps already lit keep the same voltage and the same brightness: no change. In series the voltage is shared among more lamps, so each gets less and all become dimmer.
Problem 13 Power against current squared
1. Since P = I²R, the gradient of P against I² is the resistance R.2. Gradient = 36 ÷ 4.0 = 9.0 Ω.
3. P = 9.0 × 2.5 = 22.5 W (about 22 W to 23 W from the graph).
4. I² = 18 ÷ 9.0 = 2.0 A², so I = √2.0 = 1.4 A.
Problem 14 Which bulb is brighter?
1. R = V² ÷ P. Lamp X: 144 ÷ 24 = 6.0 Ω. Lamp Y: 144 ÷ 6.0 = 24 Ω.2. R total = 30 Ω, so I = 12 ÷ 30 = 0.40 A.
3. P = I²R. X: 0.16 × 6.0 = 0.96 W. Y: 0.16 × 24 = 3.84 W.
4. Lamp Y, the lower-rated one, is brighter. In series the current is the same, so the lamp with the larger resistance gets more power.
Problem 15 Overloaded extension lead
1. Kettle: 2200 ÷ 230 = 9.57 A. Toaster: 900 ÷ 230 = 3.91 A. Lamp: 100 ÷ 230 = 0.43 A.2. Total power 3200 W, so I = 3200 ÷ 230 = 13.9 A (9.57 + 3.91 + 0.43).
3. 13.9 A is more than the 13 A rating, so the fuse will melt and cut the supply.
4. Use the kettle and toaster from separate sockets, or not at the same time, so no lead carries more than 13 A.
Problem 16 Farm lighting
1. P = 6.0 + 9.0 + 15 = 30 W.2. I = P ÷ V = 30 ÷ 12 = 2.5 A.
3. 6.0 ÷ 12 = 0.50 A; 9.0 ÷ 12 = 0.75 A; 15 ÷ 12 = 1.25 A (sum 2.5 A).
4. R = V ÷ I = 12 ÷ 2.5 = 4.8 Ω.
5. E = 0.030 kW × 5 h = 0.15 kWh; E = 30 × 18 000 s = 540 000 J.
Problem 17 Series-parallel network
1. 1/R = 1/12 + 1/4 = 4/12, so R = 3.0 Ω.2. R total = 6.0 + 3.0 = 9.0 Ω; I = 18 ÷ 9.0 = 2.0 A.
3. V across 6.0 Ω = 2.0 × 6.0 = 12 V; across the pair = 2.0 × 3.0 = 6.0 V.
4. I (12 Ω) = 6.0 ÷ 12 = 0.50 A; I (4.0 Ω) = 6.0 ÷ 4.0 = 1.5 A.
5. P = I²R or V I: 6.0 Ω: 24 W; 12 Ω: 3.0 W; 4.0 Ω: 9.0 W. Sum = 36 W, equal to the supply power 18 × 2.0 = 36 W.
Problem 18 Spot the errors
1. Resistances only add in series. In parallel 1/R = 1/5 + 1/20 = 5/20, so R = 4.0 Ω, smaller than 5 Ω.2. The formula is I = P ÷ V, not P × V: I = 1200 ÷ 230 = 5.2 A.
3. Power is I²R, not I × R. P = 3.0² × 10 = 90 W (30 would be a voltage in volts, not a power).
Problem 19 Losing power in cables
1. I = P ÷ V = 2000 ÷ 200 = 10 A.2. P(lost) = I²R = 10² × 0.50 = 50 W.
3. I = 2000 ÷ 2000 = 1.0 A.
4. P(lost) = 1.0² × 0.50 = 0.50 W.
5. The same power needs ten times less current at ten times the voltage. Since the loss is I²R, ten times less current gives 100 times less waste.
Problem 20 Design a lamp circuit
1. I = P ÷ V = 3.0 ÷ 6.0 = 0.50 A.2. R = V ÷ I = 6.0 ÷ 0.50 = 12 Ω.
3. In series R total = 24 Ω, so I = 12 ÷ 24 = 0.50 A, the rated current. Each lamp has 0.50 × 12 = 6.0 V, so both are at full brightness.
4. Energy = 12 × 2.0 = 24 W·h. Power = 6.0 W. Time = 24 ÷ 6.0 = 4.0 h.
5. The resistor must take 12 − 6.0 = 6.0 V at 0.50 A: R = 6.0 ÷ 0.50 = 12 Ω. Power wasted = I²R = 0.25 × 12 = 3.0 W.
