Question

The starter motor in a car is a small but powerful electrical device that "turns over" the gasoline-powered engine, moving it through a cycle to get it started. The starter may draw 1.80X10^2 A from the 12.0 V battery for 2.00s.
a)What is the source of the emf in this case?
b)If the time starter was running doubled, by how much would the number of electrons running in the circuit change?

Answer 1

Answer Question#61550 Physics – Electromagnetism

The starter motor in a car is a small but powerful electrical device that "turns over" the gasoline-powered engine, moving it through a cycle to get it started. The starter may draw $1.8 \cdot 10^{2}A$ from the 12.0V battery for 2.00s.

a) What is the source of the emf in this case?

b) If the time starter was running doubled, by how much would the number of electrons running in the circuit change?

Solution. a) When a voltage is generated by a battery, or by the magnetic force according to Faraday's Law, this generated voltage has been traditionally called an "electromotive force" or emf. In our case emf equal 12.0V.

b) From definition electrical current is a measure of the amount of electrical charge transferred per unit time. It represents the flow of electrons through a conductive material. $I = \frac{Q}{t} \rightarrow Q = It$. If the time starter was running doubled change charge equal to $\Delta Q = 2It - It = It$.

$\Delta Q = 1.8 \cdot 10^{2} \cdot 2 = 360C$. Charge can be represented as $\Delta Q = Ne$, where $N$ – the number of electrons, $e = 1.6 \cdot 10^{-19}C$ – the charge of the electron. Therefore $N = \frac{\Delta Q}{e} = \frac{360}{1.6 \cdot 10^{-19}} = 2.25 \cdot 10^{21}$.

Answer. a) 12V b) $N = 2.25 \cdot 10^{21}$.

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