Electrical – Why does line current become half when load becomes half of full load in a dc shunt motor

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.

Just like it's simplest to learn about a lossless inductor first, so let's start with more or less lossless motor. We'll take account of losses when we have to, but they are not essential for basic understanding.

A motor is also a generator. Spin it, and it generates volts on the armature. It doesn't matter whether it's spinning because it's a motor, or spinning because you're driving it as a generator, speed = armature_volts/k.

Pass a current through it and it generates a torque. Torque = armature_current.k

You can think of a motor as a mechanical transformer. Power in = power out. Volts x amps in = speed x torque out. When equating power, that pesky k has cancelled out. If you change the value of k, the torque constant, then the motor gets faster and delivers less torque, or vice versa, but the power balance is the same. If you run the same machine as a generator, then speed x torque in = volts x amps out.

What happens if you apply a voltage source to a motor at rest?

Two things happen, at different speeds, the first so quickly you may not notice, the second rather more slowly.

A motor armature has inductance, Larm, and resistance Rarm. At the moment of switch on, we apply V to the armature. The current starts to increase. It initially increases at such a rate that Larm x dI/dt generates a back EMF equal to the terminal voltage. The current flowing through Rarm generates a voltage IRarm which opposes V, so there is less voltage across the inductance to drive an increase of current, so the rate of current increase slows down. Eventually, the current has increased to settle at V/Rarm, with a time constant of Larm/Rarm, typically in a matter of mS.

With a 'good' motor with a low Rarm, this current will typically be very large. It's known as the 'starting' current, for obvious reasons. Small motors are rated to be started like this safely. Big motors cannot be started like this, and need some sort of soft start controller.

So far, the motor still hasn't moved, the mechanical inertia means it's not rotating yet, or has barely started. But now there is a big armature current flowing, which generates a torque, and the motor accelerates.

Once it is turning, at any speed, it generates a back EMF proportional to its speed. This back EMF reduces the effective terminal voltage available to drive current through Rarm. The armature current therefore falls (with a time constant of Larm/Rarm), and so generates less torque.

Eventually the motor reaches a balance, where it's at a speed where the generated back EMF balances off most of the input voltage, and the small difference in voltage that remains drives an armature current through Rarm, which generates enough torque in the motor to match the load torque, plus loss torques like friction and air resistance.