--- title: "Battery ignition system and spark timing calculation" subject: "Automotive Technology" chapter: 3 key: "semester_5/automotive_technology:battery-ignition-spark-numerical" source: https://www.bctengineer.com/q/semester_5/automotive_technology/battery-ignition-spark-numerical site: BCT Engineering --- # Battery ignition system and spark timing calculation Automotive Technology · Chapter 3: Ignition and Combustion - Asked in: year not given, 4+4 marks - Video: [Battery Ignition System in Hindi || Ignition system in Cars || Gear Institute](https://www.youtube.com/watch?v=s_gI6ovMqIQ) — Gear Institute Mechanical Engineering Videos and Physics by Gear Institute - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/automotive_technology/battery-ignition-spark-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-3 ## Question (a) With a neat circuit diagram, explain the working of a battery (coil) ignition system. (b) A six-cylinder, four-stroke petrol engine runs at 4000 rpm. Find the number of sparks per second and the time between sparks. If the spark advance is 24 degrees of crank angle, find the time by which the spark occurs before TDC at 4000 rpm and at 3000 rpm. The ignition coil has a primary inductance of 8 mH, breaking a primary current of 4 A, and a secondary to primary turns ratio of 120. Find the energy stored and the secondary voltage if the primary induced voltage is 280 V. ## Short answer Battery ignition system: primary current builds magnetic field; when points open, field collapse induces high voltage in secondary, distributed to spark plugs. Condenser prevents arcing. For a 6-cylinder engine at 4000 rpm: 200 sparks/s (5 ms interval); spark advance 24° gives 1.0 ms before TDC at 4000 rpm, 1.33 ms at 3000 rpm; stored energy 64 mJ; secondary voltage 33.6 kV. ## Answer #### (a) Battery (coil) ignition system ```text +-------[Ignition switch]--[Ballast R]--+ | | Battery 12V Primary winding (N1) | |--+-- Contact breaker | Secondary (N2) | + condenser | | Cam +------ ground --------------------------+ Secondary --> Distributor --> Spark plugs ``` **Working** 1. With the breaker points closed, current flows from the battery through the primary winding and builds up a magnetic field in the iron core. 2. The cam on the distributor shaft opens the points, and the primary current collapses suddenly. 3. The collapsing field induces about 250 to 300 V in the primary and, because of the turns ratio, 20 to 30 kV in the secondary. 4. The high voltage goes through the distributor to the correct spark plug, where a spark jumps across the gap. 5. The **condenser** across the points absorbs the primary induced voltage, prevents arcing at the points and makes the current collapse faster. In an electronic system, the points are replaced by a transistor switch triggered by a Hall-effect sensor or pickup coil, so there is no contact wear. #### (b) Numerical **Sparks per second:** a 4-stroke engine fires each cylinder once every 2 revolutions. $$ \text{sparks/s} = \frac{6 \times 4000}{2 \times 60} = 200 $$ Interval $= 1/200 = 5$ ms. **Time for 24 degrees advance:** crank speed at 4000 rpm $= 4000 \times 360/60 = 24\,000$ deg/s, and at 3000 rpm $= 18\,000$ deg/s. $$ t_{4000} = \frac{24}{24\,000} = 1.0\ \text{ms}, \qquad t_{3000} = \frac{24}{18\,000} = 1.33\ \text{ms} $$ For a fixed advance angle, the time available for combustion before TDC becomes shorter as speed rises. Since combustion takes roughly the same time, a centrifugal advance mechanism increases the advance angle at higher speed. **Coil energy:** $$ E = \tfrac{1}{2} L I^2 = \tfrac{1}{2}(8\times10^{-3})(4)^2 = 0.064\ \text{J} = 64\ \text{mJ} $$ **Secondary voltage:** $$ V_2 = 120 \times 280 = 33\,600\ \text{V} = 33.6\ \text{kV} $$ **Answer:** 200 sparks/s (one every 5 ms); advance time 1.0 ms at 4000 rpm and 1.33 ms at 3000 rpm; $E = 64$ mJ; $V_2 = 33.6$ kV. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the high voltage produced in a battery ignition system? A. Points close, causing a sudden rise of current B. Points open, the primary current collapses and the changing field induces high voltage in the secondary C. Condenser charges to 20 kV D. Distributor generates it Answer: B. Sudden collapse of the primary field induces 250 to 300 V in the primary and 20 to 30 kV in the secondary through the turns ratio. (Key point: Coil working) 2. What is the role of the condenser across the points? A. Stores energy for the spark B. Absorbs primary induced voltage, prevents arcing and speeds current collapse C. Distributes the spark D. Limits the battery current Answer: B. It prevents arcing at the points and makes primary current collapse faster, giving a stronger secondary voltage. (Key point: Condenser) 3. For a six-cylinder four-stroke at 4000 rpm, how many sparks per second? A. 400 B. 200 C. 100 D. 66.7 Answer: B. Each cylinder fires once per two revolutions: 6 x 4000/(2 x 60) = 200, one every 5 ms. (Key point: Sparks per second) 4. What is the time for a 24° advance at 4000 rpm? A. 1.0 ms B. 1.33 ms C. 6 ms D. 0.24 ms Answer: A. Crank speed is 24 000 deg/s, so 24° takes 1.0 ms; at 3000 rpm it is 1.33 ms. (Key point: Spark advance time) 5. What are the coil energy and secondary voltage? A. 64 mJ and 33.6 kV B. 128 mJ and 33.6 kV C. 64 mJ and 2.33 kV D. 32 mJ and 33.6 kV Answer: A. E = 1/2 L I^2 = 0.5 x 8e-3 x 16 = 64 mJ; V2 = 120 x 280 = 33.6 kV. 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