Question

A current of 5.75 A flows in the coil of wire from an applied voltage of 120 V. The motor operating at normal speed creates a back EMF of 32.0 V. What is the resistance of the coil?

Answer 1

$ϵi=V_S−IR$

$R=-[\frac{ϵi-V_S}{I}]$

Where;

ϵi=back EMF=32V

$V_s$ ==applied voltage=120v

I= current=5.75A

R=resistance

$R=-[\frac{32-120}{5.75}]=15.3ohms$

R=15.3 ohms