Question

What is the back emf of a \(120V\) motor that draws \(8.00A\) at its normal speed and \(20.0A\) when first starting?

Short answer

The back emf value is \(72\;{\rm{V}}\).

Step-by-step solution

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.

Given information and Formula to be used

\(\begin{align}{}{I_{start{\rm{ }}}} = 20\;A\\{I_{back{\rm{ }}}} = 8\;A\\V = 120\;V\end{align}\)

The current in a circuit,

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

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

Calculating the back emf

First we need find the resistance value,

\(\begin{align}{}I = \frac{V}{R}\\{I_{start{\rm{ }}}} = \frac{V}{R}\\20 = \frac{{120}}{R}\\R = 6\Omega \end{align}\)

Now finding the back emf

\(\begin{align}{}{I_{back{\rm{ }}}} = \frac{{V - {\rm{ }}Back{\rm{ }}emf{\rm{ }}}}{R}\\8 = \frac{{120 - {\rm{ }}Back{\rm{ }}emf{\rm{ }}}}{6}\\Backemf = 72\;V\end{align}\)

Therefore, the emf is obtained as \(72\;V\).