# How to find armature current in DC shunt motor in given question?

A 8.8 kW, 220 V, 1200 RPM DC shunt motor has an armature resistance of 0.1 Ω and a field resistance of 100 Ω. The rotational loss is 2200 W. Compute:
a. The rated load torque on the shaft.
b. The armature current.

I have worked out the rated load torque and the answer is 70 Nm. However, I have no idea of how to determine the armature current. I know output power is equal to developed power minus rotational loss. Hence, I got developed power as 11000 W. And developed power is equal to Ea(Ia). How to get Ea without the total current given?

• Welcome to EE.SE. Can you draw a schematic of the coil connections? (It will help you work this out.) There's a schematic button on the editor toolbar. Commented Nov 10, 2017 at 9:01

When dealing with DC motors, sketch the circuit. You know Total Power $P_T$ = 11kW and Total Voltage $V_T$ = 220V. You can work out total current.

The shunt field and armature are in parallel. You have the field resistance and the total voltage is applied across the field. Calculate field current $I_f$.

Apply KCL to determine $I_a$.

Not asked, but from there you can calculate back-emf $E_0$.

• i got 47.8A for armature current. However, the answers provided is 46.43A. Commented Nov 11, 2017 at 4:33
• It's also what I get. I agree with your torque. Commented Nov 11, 2017 at 19:27

Here is a clue. A shunt motor has two parallel current paths One through the armature the other through the shunt field. You know the field resistance and its applied voltage and hence can calculate its current and power. The field power for a given voltage remains constant and is part of the total power.

• That's what I intended the OP to see when s/he drew it out. Commented Nov 10, 2017 at 11:09
• Actually, a shunt field would be powered independently of the armature. Commented Nov 10, 2017 at 13:05
• @R Drast : In a large motor the field may well have a separate circuit from the controller but the field power is included as part of the motor rating. Otherwise could you imagine what would happen if you were sold a 9kW motor as being 90% efficient at full load - then you found out you needed another 0.5kW to energise the field. Commented Nov 10, 2017 at 23:52