Chapter 9 · 4 hours
Vehicle Electrical Systems
Practice questions
Practice questions and answers
3 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.
- Practice · 6 marks
Describe the construction and the chemical action of a lead-acid automotive battery. Explain the terms ampere-hour capacity and cold cranking amps. How is the state of charge checked?
Answer
Construction
A 12 V battery has six cells in series, each of about 2.1 V.
Terminal(+) Terminal(-)
| |
[+plates PbO2 | separator | -plates Pb] x n (cell 1)
----- cell connector ----- cell 2 ... cell 6
Container (polypropylene) filled with dilute H2SO4
- Positive plates: lead dioxide () on a lead-antimony or lead-calcium grid.
- Negative plates: spongy lead ().
- Separators: microporous plastic sheets that prevent short circuit.
- Electrolyte: dilute sulphuric acid, specific gravity about 1.26 to 1.28 when fully charged.
- Container, cell connectors, vent plugs (sealed maintenance-free types have none).
Chemical action
On discharge both plates change to lead sulphate and the acid becomes weaker (water forms). On charging, the plates return to and and the acid strengthens.
Ratings
- Ampere-hour capacity: the current a battery can supply for a given time to a specified cut-off voltage (10.5 V for 12 V) at 27 degrees C. Usually the 20-hour rate: a 60 Ah battery gives 3 A for 20 hours.
- Cold cranking amps (CCA): the current in amperes that a fully charged battery can supply for 30 seconds at -18 degrees C while keeping every cell above 1.2 V (7.2 V for 12 V). It shows starting ability.
State of charge
- Hydrometer: measures specific gravity of the electrolyte, with 1.28 = full charge, 1.20 = half, 1.12 = discharged.
- Open-circuit voltage: 12.7 V full charge, 12.2 V half.
- Load (high-rate discharge) test for a service condition.
- Practice · 4+4 marks
(a) Explain the working of an automotive alternator and charging circuit with the function of the voltage regulator. (b) Describe the starting system with its main parts, including the starter motor and solenoid.
Answer
(a) Alternator and charging system
The alternator supplies electrical loads and recharges the battery when the engine runs. It is preferred to a DC generator because it gives output even at idle, is light and has no commutator.
Rotor (field winding, slip rings) in stator (3-phase winding)
Stator -> 6-diode rectifier -> B+ (battery, loads)
Voltage regulator -> controls field current
- Rotor: claw-pole electromagnet with field winding fed through slip rings and brushes; driven by the engine belt.
- Stator: a fixed three-phase winding in a laminated core, where AC is induced as the rotor field cuts the conductors.
- Rectifier: six silicon diodes change AC to DC.
- Voltage regulator: senses the system voltage and varies the field current (by switching) to keep the output at 13.8 to 14.4 V at any speed and load. The output rises with speed otherwise, since the induced emf depends on speed and field.
- A charge-warning lamp is on when the alternator does not charge.
(b) Starting system
Battery -> Ignition switch -> Solenoid -> Starter motor
|-> shift lever -> pinion -> flywheel ring gear
- Starter motor: a series-wound DC motor, giving very high torque at low speed (up to 150 to 300 A).
- Solenoid (magnetic switch): when the key is turned, the solenoid pulls the plunger. This closes the heavy contacts to the motor and moves the shift fork that pushes the pinion (Bendix or pre-engaged drive) into the flywheel ring gear.
- Overrunning clutch: it lets the pinion transmit torque to the engine, but frees when the engine starts to prevent the armature from over-speeding.
- Reduction ratio of pinion to ring gear is about 1:10 to 1:15.
A relay and neutral safety switch may be in the circuit.
- Practice · 3+4 marks
(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.
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 on the body. It gives a neat and protected layout. Connectors are polarised and sealed. Wire size is chosen for current and voltage drop. Modern cars use a single-wire system with the body as the return (negative earth) and CAN bus to reduce the wire count.
- Fuses: a thin strip (blade fuse) that melts when current exceeds its rating (e.g. 10 A, 15 A), and protects wiring and devices from short circuits. Fusible links protect heavy circuits.
- Relays: an electromagnetic switch in which a small control current closes contacts to carry a heavy current (headlamps, horn, fuel pump), so that long, thick wires and dashboard switch contacts are avoided.
- Instrument cluster: speedometer, tachometer, fuel gauge, coolant temperature gauge, oil pressure and charge lamps. Sensors send signals; modern clusters are digital and fed by the CAN bus.
(b) Numerical
Total load
Time on battery alone
Engine running
(Charging efficiency is below 100 %, so the real time is somewhat longer.)
Answer: load W ( A); battery lasts h; charging current A; Ah is restored in h.
Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.
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