# A DC Motor is driving a constant load torque having armature resistance of 0.15 Ω and armature current of 0.25 A having connected to a voltage of 230 V. Now a resistance of 1 ohm is added into the armature circuit. The ratio of their speed will be

A DC Motor is driving a constant load torque having armature resistance of 0.15 Ω and armature current of 0.25 A having connected to a voltage of 230 V. Now a resistance of 1 ohm is added into the armature circuit. The ratio of their speed will be

#### SOLUTION

For a constant torque load, changing the armature resistance will not change the armature current. So, when the external resistance has added the speed of the motor decreases.

From the voltage equation, the back EMF of DC motor is

Eb1 = V − IaRa

= 230 − 0.25 × 60

Eb1 = 215

When the resistance of 1Ω is added in the armature circuit the new back EMF voltage

Eb2 = V − Ia(Ra + Rext)

= 240 − 60(0.25 + 1)

230 − 1.25 × 60

Eb2 = 155 V

Let the on load speed be N. As we know that back EMF of DC motor is directly proportional to the flux and speed.

Eb ∝ Nφ
Eb ∝ N (φ is constant)

The ratio of emf and speed can be equated as

$\begin{array}{l}\dfrac{{{E_{b2}}}}{{{E_{b1}}}} = \dfrac{{{N_2}}}{{{N_1}}}\\\\\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{155}}{{215}} = 0.72\end{array}$

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