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Chapter 27: Problem 6

A car battery has a \(12-\mathrm{V}\) emf, yet energy from the battery provides the \(30,000-\mathrm{V}\) spark that ignites the gasoline. How is this possible?

Short Answer

Expert verified
A car battery uses a step-up transformer to amplify its 12-Volt emf to the 30,000-Volt spark that ignites the gasoline. This is possible due to the principle of conservation of energy, as applied in transformers, where the voltage increase is countered by a proportional decrease in current, ensuring equal input and output power.

Step by step solution

01

Understanding Transformers

Transformers are devices used in electrical systems to increase or decrease the voltage of an alternating current. They work on the principle of electromagnetic induction, that is, they generate electric current in a coil as a result of changes in magnetic field in another coil.
02

How Step-up Transformers Work

In our case, the device in question is a step-up transformer. A transformer is made up of two coils, the primary and secondary coils, wound a different number of times around a toroidal core. When an AC voltage is applied to the primary coil, it creates a changing magnetic field around the core, which in turn creates a changing magnetic field around the secondary coil. Due to the varying magnetic field, an alternating current is induced in the secondary coil. If the secondary coil has more windings than the primary coil, the transformer increases the voltage and is called a 'step-up transformer'.
03

Energy Conservation

The critical principle governing a step-up transformer's operation is the principle of conservation of energy. While the voltage is increased, the current is decreased by the same factor (neglecting minimal energy losses) to ensure that the input power is equal to the output power, i.e., \(Power_{input} = Power_{output}\) or \(V_{input}I_{input} = V_{output}I_{output}\). Therefore, while the car battery delivers a voltage of 12V, the step-up transformer in the ignition coil of the car steps up this voltage to 30,000V required for the spark, consequently decreasing the current by the same factor.

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