# Effect on loading a DC motor - load torque

Could someone give a (detailed) explanation (perhaps with formulas) of how the load (torque) affects the armature current of a DC motor (separately excited or shunt or series)?

As the load torque increases the speed of the motor decreases, and assuming the terminal voltage stays constant, the EMF will become lower and thus the armature current will increase.

But could someone give me the formula that shows the relation between load torque and armature current?

Is it something like:

$T_\text{developed} = T_\text{shaft} + T_\text{friction windage} + T_\text{load}$ ,

where $T_\text{developed} = K \, \text{flux}_\text{pole} \,I_\text{armature}$ ?

But then you have

$P_d = P_s + P_{f_w} + P_l$

where $P_d = K \,\text{flux}_\text{pole}\, I_\text{armature} \,\omega_m$

In the last formula you can't see that as the increase due to $P_\text{load }$ the armature current increases because maybe someone can say the speed $\omega_m$ increases.

And how come at no-load the armature current and thus the torque developed is zero? Is it really zero, or do I have to assume there is some current flowing?

Could also explain me the principle of torque developed. Because at no load, the shaft torque is equal to torque developed, right?

• – Andy aka Dec 6 '14 at 23:20

The basic principle is simple - torque is proportional to armature current * magnetic flux. In a permanent magnet or shunt wound motor you can assume that flux is constant, so torque is just proportional to current.

However this does not take into account internal friction, windage, and magnetic losses. When the motor is running free these losses cause it to draw a no-load current (Io). Subtracting Io from total current draw leaves you with the portion that produces output torque.

Power = rotational speed x torque. As more load is applied the motor draws more current, which increases torque. However as current flows through the windings their resistance causes the effective voltage to drop, so speed decreases. Below 50% rpm the power output will also decrease, reaching zero at stall.

In a series wound motor the situation is different, because flux is not constant. With no load a series wound motor will speed up until friction and windage losses match the internal torque supplied by Io. This rpm could be very high, perhaps even high enough to destroy the motor. When a load is applied the resulting torque is proportional to current squared, because both armature and field currents contribute to magnetic force.

• So is the following correct? Power = rotational speed x Torque, since T ~ 1/wm^2 and since wm will drop due to the added load torque, it will draw more power and since the terminal voltage stays the same it will simply draw more current. Also, at no-load there is basically no developed torque (except the small E*I0 where I0 is the current you explained) – Jantje7600 Dec 7 '14 at 10:31

## protected by Community♦Oct 10 '16 at 19:48

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?