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College Physics (Urone)
Found in: Page 861
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

The motor in a toy car operates on \(6.00V\), developing a \(4.50V\) back emf at normal speed. If it draws \(3.00A\) at normal speed, what current does it draw when starting?

The current it draws when starting is \(12\;A\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of current and resistance 

Current is the term used to describe the speed at which charge moves. Resistance is the propensity of a substance to oppose the flow of charge.

Step 2: Given information and Formula to be used

\(\begin{array}{c}{I_{back{\rm{ }}}} = 3A\\V = 6V\\Backemf = 4.50\;V\end{array}\)

The current in a circuit,

\(I = \frac{V}{R},\)

where \(V\) is the potential and \(R\) is the resistance.

Step 3: Calculating the current  

First we need find the resistance value,

\(\begin{array}{c}{I_{back{\rm{ }}}} = \frac{{V - {\rm{ }}Back{\rm{ }}emf{\rm{ }}}}{R}\\3 = \frac{{6 - 4.5}}{R}\\R = 0.5\Omega \end{array}\)

Now finding current it draws when starting,

\(\begin{array}{c}I = \frac{V}{R}\\{I_{start{\rm{ }}}} = \frac{V}{R}\\{I_{start{\rm{ }}}} = \frac{6}{{0.5}}\\{I_{start{\rm{ }}}} = 12\;A\end{array}\)

Therefore, current value is \(12\;A\).

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