# Why DC Shunt Generators are Self Protective to Short Circuit?

As we know, in any kind of electrical machine there are two types of winding. One is Armature Winding which carries the load current and other is field winding which carries the DC current to create required magnetic flux. This magnetic flux serves the purpose of coupling between mechanical power and electrical power.

In DC Shunt Generators field winding is connected to the armature terminals. This means, generator voltage is applied across the terminals of Shunt field winding. This is the reason, filed winding of DC Shunt Generators are characterized by high resistance. Mind that in DC Series Machine, filed winding is connected in series with the armature winding. Therefore field winding of DC Series machine is designed to carry load current. That is why, series winding is characterized by thick wire or low resistance. Let us now discuss why DC Shunt Generators are inherently protected to short circuit faults.

Figure below shows simplified schematic drawing of DC Shunt Generator.

The equation for DC Shunt Generator can be written as

E_{a} = V_{t} + I_{a}r_{a}

But E_{a} = K_{a}ØW_{m} therefore,

K_{a}ØW_{m} = V_{t} + I_{a}r_{a}

I_{a} = (K_{a}ØW_{m} – V_{t}) / r_{a} ………………..(1)

During short circuit, the terminal voltage of generator reduces to a very low value. Ideally the terminal voltage reduces to zero during short circuit. This means V_{t} = 0. As field winding is connected across the terminal of armature, therefore the voltage across the field winding also reduces to zero. Hence, there will no more be any flow of field current I_{f} in the shunt field winding and hence no field flux Ø will be created. But there always exists some residual magnetism but this residual magnetism will be quite low to produce enough E_{a}. Thus during short circuit condition,

V_{t} =0, and I_{f} = 0

Therefore generated armature emf

E_{a} = K_{a}ØW_{m} will become ideally zero and practically will reduce to very low value.

Thus from equation (1),

Short circuit current I_{a} = (K_{a}ØW_{m} – V_{t}) / r_{a} = 0

**Thus we observe that, in DC Shunt Generator load current reduces to zero due to collapse of terminal voltage and field current. This is why, DC Shunt Generators are self-protective to short circuit fault.**