Transformers Calcuations (Just a head start)

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Mikeelliott74

Joined Jun 28, 2017
17
An ideal transformer has a primary to secondary turns ratio of 8:1. The primary current is 3 A with a supply voltage of 240 V. The transformer has iron losses of 6W and copper losses of 9W when operating on full load. Calculate the:

1 secondary voltage and current.

2. transformer efficiency at full load (pf =1)
 

crutschow

Joined Mar 14, 2008
34,285
Those calculations are for an ideal transformer.
The output current and voltage will be reduced by the iron and copper losses.
The output power thus equals the input power minus the losses.
So your calculations need to take that into account.
 

MrAl

Joined Jun 17, 2014
11,396
An ideal transformer has a primary to secondary turns ratio of 8:1. The primary current is 3 A with a supply voltage of 240 V. The transformer has iron losses of 6W and copper losses of 9W when operating on full load. Calculate the:

1 secondary voltage and current.

2. transformer efficiency at full load (pf =1)

Hi,

To get this right you probably have to model this with 4 different resistors and some voltages and currents as follows:
1. Primary applied voltage Vpa.
2. Primary series resistor Rp, which is based on part of the copper loss.
3. Primary resultant voltage Vpr due to drop in Rp.
4. Primary load resistor due to the core loss Rc.
5. Secondary series resistor Rs, which is based on part of the copper loss.
6. Secondary resultant voltage Vsr due to drop in Rs.
7. Load resistance, RL.
8. The primary current helps determine the unknowns.

So you have an applied voltage which drops due to Rp and Rc before you can apply the turns ratio, and after you apply the turns ratio you have Rs which drops more voltage which results in Vsr, then you have the load resistor.

See if you can go the rest fo the way with that. It's probably best to create an equation or set of equations to describe the complete system, then figure out how to solve for the unknowns given the input information.
 

crutschow

Joined Mar 14, 2008
34,285
Hi,

To get this right you probably have to model this with 4 different resistors and some voltages and currents as follows:
1. Primary applied voltage Vpa.
2. Primary series resistor Rp, which is based on part of the copper loss.
3. Primary resultant voltage Vpr due to drop in Rp.
4. Primary load resistor due to the core loss Rc.
5. Secondary series resistor Rs, which is based on part of the copper loss.
6. Secondary resultant voltage Vsr due to drop in Rs.
7. Load resistance, RL.
8. The primary current helps determine the unknowns.
He just needs the secondary load voltage and current so I would think you could use just one (primary or secondary) series resistor to model the copper loss and one primary parallel resistor to model the core loss.
No need to complicate the calculations with more unknowns then necessary.
 

MrAl

Joined Jun 17, 2014
11,396
He just needs the secondary load voltage and current so I would think you could use just one (primary or secondary) series resistor to model the copper loss and one primary parallel resistor to model the core loss.
No need to complicate the calculations with more unknowns then necessary.
Hi,

Yes, it does get a little complicated and i have to wonder how deep the instructor wants them to go into this.
One problem i saw was that if we reflect the secondary copper loss resistor to the primary and combine it with the primary copper loss resistor, then the voltage that actually reaches the 8:1 ideal transformer primary is reduced to a level that is unreal, because the secondary copper loss resistor comes after the core loss 'load' not before it. I would have to wonder how much difference it would make though.
Maybe it should be handled just on the basis of power too, like for example the output power is the input power minus the total loss power. Maybe we can go from there and that would probably be the simplest route.

LATER:
Looks like reflecting the secondary copper loss resistance to the primary copper loss resistance reduces the excitation current very very slightly, so slight though that the output power only decreases by much less than 1 percent, in fact, even less than 0.1 percent (something like 0.02 percent) which says it's a good estimate. In this example the core loss is very very low relative to the operating power so it doesnt matter that much for that calculation.

LATER LATER:
I have some ideal results but i want to wait to post them to give the OP time to figure this out. The interesting thing about this problem is that none of the resistor values are actually given outright, we have to find a way to calculate them all from the given info which is mostly in terms of power not resistance.
 
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