# 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

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

#### 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

Eb = V – IaRa = 0

Where,

N is the speed of the DC shunt motor

Eb is back emf

φ is flux per pole

For DC shunt motor, torque is proportional to armature current i.e. T ∝ Ia

Calculation:

Given

Total armature resistance Ra = 0.6 Ω

The full load armature current Iafull = 30 A

Terminal voltage VDC = 240 V

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

Ra = 1 + 0.6 = 1.6 Ω During Stalling torque or the breakdown torque, the speed falls to zero

Na ∝ Eb

Eb = 0 & Na = 0

According to voltage equation of motor

Eb = V − IaRa

0 = 240 − 1.6 × Ia

Ia = 150 A

∵ T ∝ φI(DC Motor)

TStalling = 150 Nm