# A d.c. shunt motor runs at 500 r.p.m. at 220 V. A resistance of 4.5 Ω is added in series with the armature for speed control. The armature resistance is 0.5 Ω. The current to stall the motor is

A d.c. shunt motor runs at 500 r.p.m. at 220 V. A resistance of 4.5 Ω is added in series with the armature for speed control. The armature resistance is 0.5 Ω. The current to stall the motor is

### Right Answer is:

44 A

#### SOLUTION

As we know that back EMF of DC motor is directly proportional to the flux and speed.

**E _{b} ∝ Nφ**

**E**

_{b}∝ N (φ is constant)We can stall the motor (i.e., reduce its speed to zero) by putting more and more load on the shaft. When the motor is stalled since N = 0 therefore E_{b} = 0

From the voltage equation, the back EMF of DC motor at no-load

**E _{b} = V − I_{a}R_{a}**

Since external resistance, R_{ext} is added in series with the motor. Hence

**E _{b} = V − I_{a}(R_{a} + R_{ext})**

0 = 220 − I_{a}(4.5 + 0.5)

I_{a} = 220 ⁄ 5

**I _{a} = 44 A**