IGCSE Physics 0625 · Section 4 · Electricity — Numerical Solutions
Electricity — Numerical Practice: Solutions
A. Charge and Current
A1
15 C in 3 s. Find current.
✓5 A
A2
2 A for 10 s. Find charge.
✓20 C
A3
50 C at 4 A. Find time.
✓12.5 s
A4
0.5 A for 2 min. Find charge.
1
Convert time to seconds
2 min = 120 s
✓60 C
A5
Conventional current vs electron flow direction.
✓Conventional current is defined to flow from positive to negative around a circuit. The actual electrons flow in the opposite direction, from negative to positive — this is a historical convention that predates the discovery of the electron.
B. E.m.f., Potential Difference and Resistance
B1
20 J moves 4 C. Find p.d.
✓5 V
B2
E.m.f. 6 V moves 3 C. Find work done.
✓18 J
B3
p.d. 12 V, current 3 A. Find resistance.
✓4 Ω
B4
R=15 Ω, I=0.4 A. Find p.d.
✓6 V
B5
R=18 Ω, V=9 V. Find current.
✓0.5 A
B6
Why does a filament lamp’s resistance rise with temperature?
✓As current flows, the filament heats up, and the increased vibration of the metal’s ions makes it harder for charge carriers to pass through, increasing resistance. On the I-V graph, this means the line curves and flattens at higher currents — each extra volt produces a smaller and smaller increase in current.
C. Series and Parallel Circuits
C1
2 Ω, 5 Ω, 3 Ω in series. Find total.
✓10 Ω
C2
8 Ω and 8 Ω in parallel. Find combined.
1
1/R = 1/R₁ + 1/R₂
1/8 + 1/8 = 2/8 = 1/4
✓4 Ω
C3
6 Ω and 12 Ω in parallel. Find combined.
1
1/R = 1/6 + 1/12
2/12 + 1/12 = 3/12 = 1/4
✓4 Ω
C4
Two 1.5 V cells in series with 5 Ω. Find current.
1
Total e.m.f. = 1.5 + 1.5 = 3 V
I = V/R = 3 ÷ 5
✓0.6 A
C5
5 A splits, one branch 2 A. Find other branch.
1
Current in = current out (Kirchhoff’s current law)
5 − 2
✓3 A
C6
Advantage of parallel lamps.
✓Each lamp receives the full supply voltage, and if one lamp fails (breaking its branch), the others stay lit since their branches are unaffected — unlike in series, where one failure breaks the whole circuit.
D. Electrical Power and Energy
D1
4 A from 12 V. Find power.
✓48 W
D2
1500 W at 230 V. Find current.
✓≈ 6.52 A
D3
100 W for 5 min. Find energy in J.
1
Convert time to seconds
5 min = 300 s
✓30000 J
D4
2 kW for 3 h. Find kWh and cost at $0.20/kWh.
1
energy (kWh) = power (kW) × time (h)
2 × 3 = 6 kWh
2
cost = energy × price
6 × 0.20
✓6 kWh; cost = $1.20
D5
5000 J in 25 s. Find power.
✓200 W
D6
Joule vs kilowatt-hour.
✓Both are units of energy. A joule is the energy transferred by 1 watt of power in 1 second — a very small amount. A kilowatt-hour is the energy transferred by 1000 W (1 kW) of power running for 1 hour (3600 s) — a much larger amount, used for billing because household energy use over an hour would otherwise be an enormous number of joules.
E. Transformers and Power Transmission
E1
Np=100, Ns=500, Vp=20 V. Find Vs, type.
1
Vp/Vs = Np/Ns → Vs = Vp × (Ns/Np)
20 × (500/100)
✓Vs = 100 V — this is a step-up transformer (Vs > Vp)
E2
Vp=230 V, Vs=11.5 V. Find Np:Ns, type.
1
Np/Ns = Vp/Vs
230 ÷ 11.5 = 20
✓Np:Ns = 20:1 — this is a step-down transformer (Vs < Vp)
E3
Ip=2A, Vp=230V, Vs=4600V. Find Is.
1
IpVp = IsVs
2 × 230 = Is × 4600
✓0.1 A
E4
Np=1000, Ns=100, Ip=0.5A. Find Is.
1
Combining Vp/Vs=Np/Ns with IpVp=IsVs gives: Is = Ip × (Np/Ns)
0.5 × (1000/100)
✓5 A
E5
R=4 Ω, I=2 A. Find power lost.
✓16 W
E6
10000 W at 1000V(10A) vs 10000V(1A), R=5Ω. Compare losses.
1
At 1000 V: P = I²R
10² × 5 = 500 W
2
At 10000 V: P = I²R
1² × 5 = 5 W
✓500 W lost at 1000 V vs only 5 W lost at 10000 V — a 100× reduction from a 10× increase in voltage
E7
Why transmit at high voltage?
✓For a fixed amount of power delivered, a higher transmission voltage requires a correspondingly lower current. Since cable power loss (P = I²R) depends on the SQUARE of the current, even a modest reduction in current produces a much larger reduction in wasted power — making high-voltage transmission far more efficient.
F. Potential Dividers
F1
20 Ω and 30 Ω in series across 10 V. Find voltage across 30 Ω.
1
The two resistors share the 10 V in the ratio of their resistances; the 30 Ω share is
V₂ = 10 × (30/(20+30))
✓6 V
F2
40 Ω and 60 Ω in series across 20 V. Find voltage across 60 Ω.
1
V₂ = V_total × R₂/(R₁+R₂)
20 × (60/(40+60))
✓12 V
F3
15 Ω has 5 V; 25 Ω in series. Find voltage across 25 Ω.
1
R₁/R₂ = V₁/V₂ → rearrange for V₂
15/25 = 5/V₂ → V₂ = (5 × 25)/15
✓≈ 8.33 V
F4
Purpose of a potential divider.
✓A potential divider splits a supply voltage between two (or more) resistors in the ratio of their resistances, producing a smaller, chosen voltage from a larger supply — useful for supplying the correct, often variable, voltage to another part of a circuit.
Mastery Test — Answer Key
- 5 A (40 ÷ 8)
- 45 C (3 × 15)
- 5 V (60 ÷ 12)
- 8 Ω (20 ÷ 2.5)
- 20 Ω (4 + 7 + 9)
- 5 Ω — 1/R = 1/10+1/10 = 2/10 → R = 5
- 1320 W (6 × 220)
- 10000 J (250 × 40)
- 24 V, step-down — Vs = 120 × (60/300) = 24 V; Vs < Vp
- 50 W — P = I²R = 5² × 2
