Question

The armature windings of a dc motor have a resistance of 5.0 Ω. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. The load is increased so it causes the motor to run at half speed.
What will be the current in the motor in this case?

Answer 1

Answer:

The current in the motor in this case is 13.2 A

Explanation:

Given:

Resistance $R =$ 5 Ω

Emf $E = 120$ V

Induced emf $E _{induced} = 108$ V

When motor run at half speed due to load increased then induced emf is also reduced to half of its value

So new induced emf in our case is given by,

$E_{induced } = \frac{108}{2} = 54$ V

  $I = \frac{V}{R}$

Where $V =$ $E - E_{Induced }$

  $I = \frac{120-54}{5}$

  $I = 13.2$ A

Therefore, the current in the motor in this case is 13.2 A