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Battery runtime calculation and lighting circuit

Short answer

Wiring harness bundles colour-coded wires; fuses melt on overcurrent; relays switch heavy loads with low control current; instrument cluster shows speed, fuel, temp. Total load 190 W (15.83 A). Usable capacity 48 Ah, battery lasts 3.03 h. Alternator output 30 A, so charging current 14.17 A; restoring20 Ah takes 1.41 h.

(a) Write short notes on the wiring harness, fuses and relays of a vehicle, and the instrument cluster. (b) A vehicle with a 12 V, 60 Ah battery has two 55 W headlamps, two 5 W tail lamps, 10 W of instrument lighting and a 60 W wiper motor switched on with the engine stopped. Taking 80 % of the battery capacity as usable, find the time for which the battery can supply this load. When the engine is running, the alternator gives 30 A; find the current available to charge the battery and the time to restore 20 Ah.

(a) Wiring harness, fuses, relays, instruments

  • Wiring harness: a bundle of colour-coded, insulated copper wires of different gauges, tied together with tape and sleeves and clipped to the body. It gives a neat, protected layout. Connectors are polarised and sealed. Wire size is chosen for the current and the allowed voltage drop. Cars use a single-wire system with the body as the return path (negative earth), and modern cars use the CAN bus to reduce the number of wires.
  • Fuses: a thin strip (as in a blade fuse) that melts when the current exceeds its rating (e.g. 10 A, 15 A), protecting wiring and devices from short circuits. Fusible links protect heavy circuits.
  • Relays: electromagnetic switches in which a small control current closes contacts that carry a heavy current (headlamps, horn, fuel pump). This avoids long, thick wires to the dashboard switch and protects the switch contacts.
  • Instrument cluster: speedometer, tachometer, fuel gauge, coolant temperature gauge, and oil pressure and charge warning lamps. Sensors send the signals; modern clusters are digital and fed by the CAN bus.

(b) Numerical

Total load

P=2(55)+2(5)+10+60=190 W,I=PV=19012=15.83 AP = 2(55) + 2(5) + 10 + 60 = 190\ \text{W}, \qquad I = \frac{P}{V} = \frac{190}{12} = 15.83\ \text{A}

Time on battery alone

usable capacity=0.8×60=48 Ah,t=4815.83=3.03 h\text{usable capacity} = 0.8 \times 60 = 48\ \text{Ah}, \qquad t = \frac{48}{15.83} = 3.03\ \text{h}

Engine running (same load switched on)

Icharge=30−15.83=14.17 A,t=2014.17=1.41 hI_{charge} = 30 - 15.83 = 14.17\ \text{A}, \qquad t = \frac{20}{14.17} = 1.41\ \text{h}

(Charging efficiency is below 100 %, so the real time is somewhat longer.)

Answer: load =190= 190 W (15.8315.83 A); the battery lasts 3.033.03 h; charging current =14.17= 14.17 A; 2020 Ah is restored in ≈1.41\approx 1.41 h.

✓ Answer checked by Claude Opus 5.5 (high effort)

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