The resistance of the armature in the motor shown in Fig. $33-2$ is $2.30 \Omega$. It draws a current of $1.60$ A when operating on $120 \mathrm{~V}$. What is its back emf under these circumstances?
The motor acts like a back emf in series with an $I R$ drop through its internal resistance. Therefore,
or
$$
\begin{array}{l}
\text { Line voltage }=\text { back emf }+I r \\
\qquad \text { Back emf }=120 \mathrm{~V}-(1.60 \mathrm{~A})(2.30 \Omega)=116 \mathrm{~V}
\end{array}
$$