Question

Why the magnitude of field current in dc shunt generator is lesser as compared to armature current?

Answer 1

Answer:

Explanation:

Let us take a typical DC shunt motor

Let the supply voltage be V and let the armature resistance be Ra. Let the armature current be Ia and back emf be E. By applying basic KVL,

V-IaRa=E

We also know that E= k * flux * speed where k is the machine constant

Putting that back in the first equation, we get

V-IaRa=k*flux*speed

We know that flux is produced by current flowing through the field winding. Therefore, flux produced is directly proportional to the field current i.e flux = k1 * If. Plugging that back in the previous equation, we get

V-IaRa= k * k1* If * speed

From this, we get speed= (V-IaRa)/(k*k1*If)

So, if we keep the supply voltage and the load constant (by extension, Ia constant), we find that the motor speed is inversely proportional to the field current.

EDIT 1:

I have a better answer to it, a more detailed explanation from Chapman’s book of Electrical Machines.

Once you reduce the field current, you reduce the magnetic flux being fed to the machine

Once the magnetic flux fed is reduced, the back EMF (Kϕw) reduces.

Once that reduces, the armature current increases by the equation (Vt-Kb)/R=I

Once I increases, the developed torque increases by the equation Td=KϕI

Once the developed torque increases over the load torque, the speed of the motor will automatically increase( Think about it, logical)

Once w increases, back EMF increases, decreasing I

As I reduces, the motor settles to a new value of developed torque

Answer 2

Let us take a typical DC shunt motor

Let the supply voltage be V and let the armature resistance be Ra. Let the armature current be Ia and back emf be E. By applying basic KVL,

V-IaRa=E

We also know that E= k * flux * speed where k is the machine constant

Putting that back in the first equation, we get

V-IaRa=k*flux*speed

We know that flux is produced by current flowing through the field winding. Therefore, flux produced is directly proportional to the field current i.e flux = k1 * If. Plugging that back in the previous equation, we get

V-IaRa= k * k1* If * speed

From this, we get speed= (V-IaRa)/(k*k1*If)

So, if we keep the supply voltage and the load constant (by extension, Ia constant), we find that the motor speed is inversely proportional to the field current.

EDIT 1:

I have a better answer to it, a more detailed explanation from Chapman’s book of Electrical Machines.

Once you reduce the field current, you reduce the magnetic flux being fed to the machine

Once the magnetic flux fed is reduced, the back EMF (Kϕw) reduces.

Once that reduces, the armature current increases by the equation (Vt-Kb)/R=I

Once I increases, the developed torque increases by the equation Td=KϕI

Once the developed torque increases over the load torque, the speed of the motor will automatically increase( Think about it, logical)

Once w increases, back EMF increases, decreasing I

As I reduces, the motor settles to a new value of developed torque