# A 240 V DC shunt motor has an armature resistance of 0.6 Ω. The full load armature current is 30 A. The ratio of the stalling torque to the full load torque when the resistance of 1 Ω is connected in series with the armature is

### Right Answer is:

5

#### SOLUTION

Stalling torque is torque at which speed falls to zero, i.e

N = 0

For DC shunt motor, speed is proportional to

$N \propto \frac{{{E_b}}}{\varphi }$

For stalling torque

E_{b} = V – I_{a}R_{a} = 0

Where,

N is the speed of the DC shunt motor

E_{b} is back emf

φ is flux per pole

For DC shunt motor, torque is proportional to armature current i.e. T ∝ I_{a}

**Calculation:**

**Given**

Total armature resistance R_{a} = 0.6 Ω

The full load armature current I_{afull} = 30 A

Terminal voltage V_{DC} = 240 V

Now 1Ω resistance is connected in series with armature then total resistance become

R_{a} = 1 + 0.6 = 1.6 Ω

During Stalling torque or the breakdown torque, the speed falls to zero

N_{a} ∝ E_{b}/φ

E_{b} = 0 & N_{a} = 0

According to voltage equation of motor

E_{b} = V − I_{a}R_{a}

0 = 240 − 1.6 × I_{a}

I_{a }= 150 A

∵ T ∝ φI_{a }(DC Motor)

T_{Stalling} = 150 Nm

T_{Full-load} = 30 Nm

The ratio of stalling to full load torque is

T_{Stalling}/T_{Full-load } = 150/30

**T _{Stalling}/T_{Full-load } = 5**