Chapter 21, Problem 23P

To determine

The amount of back emf of a motor.

Answer to Problem 23P

Solution:

Back emf of motor = $\text{90}\text{.07 V}$ .

Explanation of Solution

Given info:

  $\begin{array}{l}\begin{array}{l} \text{Applied emf}\left(E_A\right)=\text{120 V}\end{array}\hfill \end{array}$

   $\begin{array}{c} \text{Resistance}\left(R\right)=\text{3}\text{.65 }\Omega \end{array}$

     $\begin{array}{c} \text{Current}\left(I\right)=\text{8}\text{.20 A}\hfill \end{array}$

A motor will have a back emf which opposes the applied emf (supply voltage), as the motor speeds up, the back emf increases and it affects the difference between $E_A$ (supply voltage) and back emf which makes the current to flow through the armature resistance.

Formula to be used:

By Ohm’s law,

  $V=IR$

Calculation:

Applying values,

  $\begin{array}{l}V=\left(8.20\right)\left(3.65\right)\end{array}$

  $\begin{array}{c}\text{ =29}\text{.93 V}\end{array}$

Therefore,

  $\begin{array}{l}{\text{E}}_A=E_{Back}+V\end{array}$

$\begin{array}{c}\Rightarrow E_{Back}=E_A-V\end{array}$

$\begin{array}{c}\text{ =120-29}\text{.93}\end{array}$

$\begin{array}{c}\text{ =90}\text{.07 V}\end{array}$

Conclusion:

Therefore, the back emf of the motor is $\text{90}\text{.07 V}$ .