---
title: "Battery runtime calculation and lighting circuit"
subject: "Automotive Technology"
chapter: 9
key: "semester_5/automotive_technology:battery-load-wiring-numerical"
source: https://www.bctengineer.com/q/semester_5/automotive_technology/battery-load-wiring-numerical
site: BCT Engineering
---

# Battery runtime calculation and lighting circuit

Automotive Technology · Chapter 9: Vehicle Electrical Systems

- Asked in: year not given, 3+4 marks
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- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/automotive_technology/battery-load-wiring-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-9

## Question

(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.

## 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.

## Answer

#### (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**

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

**Time on battery alone**

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

**Engine running** (same load switched on)

$$
I_{charge} = 30 - 15.83 = 14.17\ \text{A}, \qquad t = \frac{20}{14.17} = 1.41\ \text{h}
$$

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

**Answer:** load $= 190$ W ($15.83$ A); the battery lasts $3.03$ h; charging current $= 14.17$ A; $20$ Ah is restored in $\approx 1.41$ h.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why are relays used for headlamps and horn?
   A. A small control current switches a heavy current, avoiding long thick wires via dashboard switches
   B. They convert AC to DC
   C. They protect against short circuit by melting
   D. They store energy
   Answer: A. A relay is an electromagnetic switch; fuses are the parts that melt. (Key point: Relay)

2. What is the total load current?
   A. 15.83 A (190 W at 12 V)
   B. 23.3 A
   C. 190 A
   D. 8.5 A
   Answer: A. P = 110 + 10 + 10 + 60 = 190 W; I = 190/12 = 15.83 A. (Key point: Total load)

3. How long can the battery supply the load, using 80 % of 60 Ah?
   A. 1.41 h
   B. 3.79 min
   C. 3.79 h
   D. 3.03 h
   Answer: D. Usable 48 Ah / 15.83 A = 3.03 h. (Key point: Battery time)

4. With a 30 A alternator and this load running, how long to restore 20 Ah?
   A. 1.26 h
   B. 3.03 h
   C. 0.67 h
   D. 1.41 h
   Answer: D. Charge current = 30 - 15.83 = 14.17 A; 20/14.17 = 1.41 h. (Key point: Charging)

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