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Q42PE

Expert-verifiedFound in: Page 861

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 \(12\;A\).

**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.**

** **

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

The current in a circuit,

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

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

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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