Transformers II (1 Viewer)

[Suvat]

Hey guys, I've got another problem with transformers
We performed a prac to detemine the efficiency of a transformer and used the formula
Measured Voltage in secondary coil
Efficiency = --------------------------------------------- x 100%
Theoretical Voltage in secondary coil
where the theoretical voltage was obtained through the formula
N(p) V(p)
----- = ------ where V(s) was the theoritical voltage
N(s) V(s)

However, I am unsure of the validity of the first formula as I had always thought that the formula for efficiency was:
Power out
----------------------------------------- x 100%
Power in (or theoritical power out)

and since P=VI, I wondered whether it's valid to defer consideration of the currents.

I would also like to know whether the power disspated in transformers (since no transformer is 100% efficient) eats into the current output, voltage output or both
eg, if a transformer has 10 turns in the primary coil and 20 turns in the secondary coil and a current of 10 amps carrying 10 volts is fed into the primary coil it is theoretically supposed to generate an output of 20V and 5A in the secondary coil. But since some power is lost, does it come in the form of less than 20V or less than 5A or both? For example if the transformer was 50% efficient what will be the voltage and current induced in the secondary coil?

Thanks.

 

[kini mini]

Originally posted by Suvat

I am unsure of the validity of the first formula as I had always thought that the formula for efficiency was:
Power out
----------------------------------------- x 100%
Power in (or theoritical power out)

and since P=VI, I wondered whether it's valid to defer consideration of the currents.

I think that in considering the voltage induced, you are indirectly considering the current - after all, it's the potential difference (voltage) that creates the current. You're not really deferring consideration of the currents.

I would also like to know whether the power disspated in transformers (since no transformer is 100% efficient) eats into the current output, voltage output or both

Good question, I've never really thought about this. Starting once more from the general principle (Lenz) of inducing a current, it seems to me that the current induced will be less than the theoretical value (usual reasons, resistance in the windings leading to loss as heat, reasons to do with the magnet itself which I've forgotten...), and therefore the voltage will be lower - so both.

 

[Akira_Tikira]

Hi Let me comment on a statement by the person before me whatever your name was....

It is a fundamental mistake to make....A potential difference is created so current can flow.....what comes first Potential difference then curren flows....

Its like saying a pressure in the pump is required before a current will flow, So we talk about infuced emf and induced current, obviously induced emf comes first.

Also interesting, a bird standing on the powerline will not get electrocuted, that is because the potential difference is the same, so a current will not flow.

Thanks, Just make sure you guys get the fundamentals right

 

[kini mini]

oops, you're quite right Akira - serves me right for posting late at night and getting one phrase in front of another.

 

[wogboy]

The inefficiency of a transformer reduces both its output voltage and output current by the same factor (**note that the "output current" of a transformer really means the MAXIMUM current that can flow in the secondary circuit because the current given out by a transformer varies depending on how much power is drawn from the secondary circuit).

To calculate the efficiency of a transformer, dividing the measured voltage ratio by the number of turns ratio (which is the theoretically correct voltage ratio) won't give the correct efficiency. What you need to do is to square this value, and this will be the value of the efficiency (e), so the formula is (where vs = secondary voltage, vp= primary voltage, ns = secondary turns, np = primary turns):

e = [(vs/vp) / (ns/np)] ^2,

alternatively, e = [(vs/vp) * (np/ns)] ^2,

not simply e = (vs/vp) / (ns/np)

Similarly, if you're given the efficiency of a transformer, it's primary voltage & current, and no of turns, then you can work out the output voltage and the output current (remember **), according to the following formulae:

vs = vp * (ns/np) * sqrt(e)

and

is = ip * (np/ns) * sqrt(e)

If you want to verify this, multiply the two expressions, vs & ip, and you get:

vs* is = vp * ip * [sqrt(e) * sqrt(e)] * (np / ns) * (ns / np)
= (vp * ip) * e
= vp * ip * e

so,

(power out) = (efficiency) * (power in)

which of course makes sense.

So to answer suvat's question:

vp = 10V
e = 50% = 0.5

so vs = vp * sqrt(e) * (ns/np)
= 10 * sqrt(0.5) * 2
= 14.14 V (as opposed to the ideal voltage of 20V)

& is = ip * sqrt(e) * (np/ns)
= 10 * sqrt(0.5) * 0.5
= 3.54 A (as opposed to the ideal current of 5A)

(Don't worry too much if you don't fully understand what's going on and why, I'm sure you won't need to know all this for the HSC level Physics)