# Power consumption of Inductive Capacitive load?

I came across the concept of Apparent Reactive and Real power. It says ideal inductive and capacitive circuits do not consume any power. So my question is - Do inductive motors consume any power at all? Of course they do .. Then how is that true?

So my question is does inductive motors

I think you mean induction motor.

Electric motors are energy conversion devices and an ideal electric motor converts 100% of the electrical power it receives from the circuit to mechanical power - the ideal motor itself does not consume power but, rather, changes its form.

Of course, for real motors, there are loses due to e.g., winding resistance, friction, etc.

Now an inductor and a capacitor are energy storage circuit elements - they alternately store and then release energy from and to the electric circuit to which they are attached.

So, for ideal inductors and capacitors, there is no loss - on average, their associated electric power is zero. The instantaneous power associated with an ideal reactive element is alternately positive and negative.

However, for a resistance, the electric power from the attached circuit is dissipated - converted to heat. The instantaneous power is always positive.

That's "induction motor", not "inductive motor" — two different concepts.

An induction motor uses magnetic induction to get current flowing in the rotor instead of using direct electrical contact through brushes. In that sense, it works somewhat like a transformer.

Unloaded motors do indeed behave mostly like inductors in the electrical sense, but any mechanical load on them appears as a resistive electrical load in parallel with that inductance, and that's where the power consumption occurs.

• So when motor is running freely (No load) what will be the power consumption roughly ? Commented May 14, 2014 at 11:40
• There will be some electrical loss (resistance in the windings, etc.) plus some mechanical losses (bearing friction, air resistance, etc.), and these represent the power consumption of an "unloaded" motor. In reality, there's always some load. Commented May 14, 2014 at 11:49

All electrical devices draw power, there will definitely be power drawn from your supply.

If you look at a DC motor, then your power will be equal to P = IR, which will be the plate rating of the motor, normally measured in kW. Things get more complicated when you look when voltage is changing.

On a DC motor, when you switch it on, you apply voltage to the terminals, so the voltage changes from 0V to xV, where x is the voltage of your source. In between the motor being switched and stead state, the voltage will allow current to flow, but as a motor requires extra current to get the state state set up, it will draw more current than the plate rating of the motor. This current is called the inrush current, or starting current. This extra current will be required for a few milliseconds while the motor gets itself to steady state. As the cables that connect the source to the motor and the wires that make the windings within the motor itself have their own associated resistance, capacitance and inductance, it is important to consider what effect they have on the complete system. Resistance has the same effect during start up (dynamic) or during steady state but induction and capacitance only have effect during dynamic state, thus your startup current will have overcome these factors during the starting (or dynamic conditions) of you motor. This is because they themselves reach a steady state under static DC voltages and currents.

When you look at AC power, the system is always in dynamic mode as the voltage is not static. So when you look at AC motors, you will see something called power factor.

When a dynamic voltage is applied to a load, the current is not aligned with the voltage. That means that either the current leads the voltage or lags the voltage. The easiest way to see this is with a vector diagram. Resistance is from the origin to the right, induction is from the origin straight up and capacitance is from the origin straight down.

Now if you take a DC motor, let's say a 10VDC 10kW motor, and drew it on the drawing, you would put a marker on the resistance axis for the resistance you knew the motor had, so this example would be at 1ohm.

An AC motor, and the cables connecting it to the source, has a intrinsic resistance, inductance and capacitance. All 3 effect the amount of power required from the source to make the motor spin and the voltage is dynamic. This is where power factor comes in.

The work done by the motor is equivalent to the kW rating of the motor, but only the power dissipated within the resistance of the motor does the work (real power). Any power dissipated in the capacitance or inductance of the motor or cables is wasted (apparent). Luckily, capacitance and inductance directly cancel each other out, so if you have a motor with a natural inductance of x, you simply wire a capacitor, also with capacitance x, parallel to it and hey presto, you have cancelled the wasted power out and all you power drawn will go to spinning the motor.

Don't for get that your source also sees the total power required, i.e. the real and apparent combined, which means by reducing the apparent power of each load, you can add more loads to the same source.

In reality we aim to reduce the power factor to between 0.95 and 0.98 for a couple of reasons:

1. A "worse" powerfactor than 0.95 is just wasting energy and money
2. A power factor less than 0.95 is normally fined by the utility supplier
3. getting power factor better than 0.98 is really expensive and not worth the investment
4. hitting unity power factor (pf = 1) causes other issues with harmonics pollution of the supply

Because your energy bill is reduced by installing power factor correction equipment, the equipment will pay for itself in about 3 years normally, so you pretty much always have them.

If you want more info, let me know as this is a tricky topic to get your head round and involves intermediate knowledge (degree year 1/2 level knowledge to fully understand).