Question

the back emf in a motor is 66 v when operating at 1300 rpm
part a
what would be the back emf at 3000 rpm if the magnetic field is unchanged?
express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Use the formula for back - emf

The back - emf \(E\) of a motor is given by \(E = k\omega\), where \(k\) is a constant (since the magnetic field is unchanged, \(k\) remains the same) and \(\omega\) is the angular velocity. If we assume a linear relationship (proportionality) between \(E\) and the rotational speed \(n\) (in rpm, as \(\omega\propto n\) for a simple motor model), we can use the proportion \(\frac{E_1}{n_1}=\frac{E_2}{n_2}\).

We are given \(E_1 = 66\ V\), \(n_1=1300\ rpm\), and \(n_2 = 3000\ rpm\).

Step2: Solve for \(E_2\)

From \(\frac{E_1}{n_1}=\frac{E_2}{n_2}\), we can cross - multiply to get \(E_2=\frac{E_1\times n_2}{n_1}\).

Substitute the values: \(E_2=\frac{66\times3000}{1300}\)

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Rounding to two significant figures, \(E_2 = 150\ V\)

Answer:

\(150\ V\)