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 V . What is its back emf under these circumstances?
The motor acts like a back emf in series with an IR drop through its internal resistance. Therefore,
Line voltage $=$ back emf $+I r$
or
Back cmf $=120 \mathrm{~V}-(1.60 \mathrm{~A})(2.30 \Omega)=116 \mathrm{~V}$