1. Jun 7, 2010

### salman213

1. Lets say I put a resistor with a value 1 ohm on the secondary side..

what happens???

[PLAIN]http://img256.imageshack.us/img256/9639/67417248.jpg [Broken]

The thing is, in theory I just calculate the Z value and find out what it is. BUT, why does it have to be that. Practically, what would happen if I take that Z and replace it wih 1 ohm.

Note: I am not actually solving for Z because this is not really a hwk question I jsut wanted to know what would happen if I place 1 ohm resistor on the load side.

Last edited by a moderator: May 4, 2017
2. Jun 8, 2010

### n.karthick

As you mentioned, the transformer rating is 2400W. It means that the maximum load it can handle is Zmax=6 ohm. Now if you connect Z=1ohm, you are actually overloading the transformer, since now Z<Zmax. So the current drawn in sec. side is 120 A and power P=120*120=14.4kW. The current in the primary side can now be calculated as you know Vp and P. (Note: The transformer is assumed to be loss-less)

3. Jun 8, 2010

### salman213

That wording is a bit confusing since you are saying you are overloading the transformer, yet I am applying LESS load on the secondary side. I guess when someone says overloading the transformer, you have to look at how much power is needed in the secondary side.

So my question is, if I overload the transformer (by putting Z<Zmax.. which will require higher current), what will happen?

only 2400 W can be provided by the primary but the secondary needs 14.4 kW. What will the transformer do? What will current be in the secondary side?

THANKS! :)

4. Jun 9, 2010

### n.karthick

No! you are overloading the transformer by connecting 1 ohm. I typed wrongly as Zmax sorry. As per transformer rating 6 ohm is the minimum Z that can be connected, so Zmin=6 ohm. If Z< Zmin transformer is overloaded, because as you mentioned power drawn in the secondary is more than the rating.

The power on primary side is not fixed. Whatever power is drawn by load in secondary side will be supplied by primary. So primary will also deliver 14.4kW (plus the losses in transformer). (This will not happen in a practical case, since overloaded transformer will be heated up and its windings may be burnt) As you know transformer is a passive device, the primary will respond to load demand on secondary and in this case primary current will be 60 A (i.e more than the rated value.) The secondary current will be 120 A.

5. Aug 24, 2010

### zgozvrm

You're both right; it's a matter of perspective.

Generally, people think of the resistance as the load, not the power or current. Therefore, a bigger load would mean a higher resistance and a 1 ohm resistance would be under-loading the transformer, bringing the load closer to short-circuit (low resistance / high amperage).

Transformers will draw as much power from the primary as they need to supply the load on the secondary. So, in this case, a 1 ohm load on the secondary would require 14.4 kW which is more power than the transformer can handle, causing it to do more work than it is designed to do, thus "overloading" the transformer.

Perhaps, we should consider the transformer underloaded (too small of a load) and overworked (too much power required from it).

6. Aug 24, 2010

### ms_kit

I am confused about this. Say the rating is 2400 W and now the primary delivers 14.4kW because secondary coil is now loaded and requires 14.4kW.
Where does this "extra" power of 14.4 - 2.4 kW come from?

7. Aug 24, 2010

### Zryn

Theoretically you have an unlimited power source.

Realistically, from a GPO you would trip a circuit breaker and blow fuses and maybe start a fire aswell if youre lucky. You might damage the house wiring, the transformer, the load, and anything else. Its all based on the amount of time and the ratings for each component. Things are usually designed for massive spike voltages and currents over very minor time periods. So maybe nothing major, or maybe it could get exciting! Knowing the power rating of the transformer would translate into a more mathematical answer.

Realistically, from a laboratory power source with current and/or voltage limiting, you would just hit the upper limits for whichever variable you have set, and Ohms Law would sort out the other one. Probably a red LED or two would come on somewhere to indicate youre doing something bad.

8. Aug 25, 2010

### ms_kit

Unlimited power source???
Usually the A.C. supply at the primary coil is from an A.C. generator at power station. hence according to conservation of energy (power).. kinetic energy is converted to electrical energy.

With a load ( e.g. 1 ohm) at the secondary coil and hence drawing power of 14.4 kW (instead of the no load scenario of 2400 W)
how does this conservation of energy apply throughout this power transmission system?

Something must be compromised.

9. Aug 25, 2010

### Zryn

What power transmission system are you talking about? We're talking about the basic transformer theory of conservation of energy on a peice of paper.

If you have a secondary power of 14.4kW then you MUST have a primary power of at least 14.4kW, but why does the source of this theoretical power matter? If it is a theoretical AC generator or a theoretical unlimited power source it makes no difference to the question.

If you're after what happens realistically, then I would refer you to the rest of my previous post.

The current you're drawing with such a low load will activate protection mechanisms, depending on where youre drawing it from. In a house this will be circuit breakers and/or fuses and in a lab this will be current limiting mechanisms in your power supply, and then circuit breakers and/or fuses.

If you're asking for numbers, then they can't be given unless you know the transformer power characteristics. If its a 2.4kVA (2.4kW) transformer that you're trying to put 14.4kW of power through, something in the protection circuitry will activate or the wires and components somewhere throughout your system will melt and/or burn.