Electromagnetic induction > The motion of conductors in a constant magnetic field. Electric motors > Problem 11.1.28

Statement

$11.1.28.$ The rotor frequency of a DC motor connected to a battery circuit with an EMF of 24 V and a total resistance of 20 $\Omega$ is 600 rpm at a current of 0.2 A. What EMF will the same motor develop when operating as a dynamo (generator) at a frequency of 1200 rpm?

Solution

It is known that the induced EMF of an electric motor is directly proportional to the magnetic flux and its frequency. Since only the motor's frequency varies, we could write the following:

\begin{equation}
\mathcal{E}_i=k\omega
\end{equation}

where $\mathcal{E}_i$ is the induced EMF, $\omega$ is angular frequency of the motor and $k$ is a constant.

For the first case we have:

\begin{equation}
\mathcal{E}_1+k\omega_1=IR
\end{equation}

where $ \mathcal{E}_1=24V $, $I=0.2$A,$R=20$$\Omega$ and $\omega_1=600$ rpm

Similarly,for the second case we could write:

\begin{equation}
\mathcal{E}_2+k\omega_2=0
\end{equation}

where $\omega_2=1200$ rpm

By substituting $k$ from eq(2) to eq(3) we get the answer for $\mathcal{E}_2$:

\begin{equation}
\mathcal{E}_2=(\mathcal{E}_1-IR)\frac{\omega_2}{\omega_1}=40V
\end{equation}

Answer

$\mathcal{E}_2=40V$