Suppose we have a 100 V A.C. supply, and we connect a 1 ohm resistor with it. Now the current through the resistor will be 100 A. Now we connect a 1:2 step up transformer with the A.C supply, and then connect the 1 ohm resistor with the secondary. Now, voltage across the secondary will be 200 V. And assuming that primary and secondary coils have no resistance, the current through the resistor will be 200 A. And the current through the primary will be 400 A. So, current has increased from 100 A to 200 A. But, step up transformers are used to reduce current to decrease $I²R$ loss through transmission lines. How is it?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Please add a circuit diagram. Any question about circuits should have a diagram. $\endgroup$– DanielSankCommented Nov 5, 2023 at 16:03
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
But, step up transformers are used to reduce current to decrease I²R loss through transmission lines. How is it?
The load on a transmission line doesn't act like a fixed 1 ohm resistor.
The load on a power transmission line is, for example, a step-down transformer whose secondary is connected to a neighborhood of houses that consume a fixed amount of power (well, not a "fixed" amount, but an amount that's controlled by the occupants of the houses and not by the engineer designing the transmission line).
So when you increase the voltage on the line, the current required goes down rather than up.
And the $I^2 R$ loss that we reduce is the loss in the series resistance of the line itself, not the loss in the load.
answered Nov 5, 2023 at 16:08
-
$\begingroup$ So, you mean that current supplied to the primary remains same but the current in the secondary reduces if we put in a step up transformer between source and load? $\endgroup$ Commented Nov 5, 2023 at 16:15
-
$\begingroup$ There are two transformers. At the input end you step up from there generator voltage to a high voltage for transmission. At the other end you step down to three voltage required by the load. $\endgroup$ Commented Nov 5, 2023 at 17:49
$\begingroup$
$\endgroup$
2
If you are stepping up the voltage on the source side, you also need to step it down on the load side (if you aren't adjusting your load accordingly), as The Photon has also mentioned.
A more suitable toy problem is a step-up transformer ($1$:$N$) and step-down transformer ($N$:$1$) connected via a resistor which models transmission losses, with a fixed load connected across the secondary of the step-down transformer. Then, it can easily be shown that the higher the step-up/down ratio $N$, the lower the transmission loss.
-
$\begingroup$ So, suppose we have a 100 V A.C supply, and a 1000 W load. If connected without transformer, it draws 10 A current. Now we first connect a 1:10 step up transformer and then a 1:10 step down transformer. Now the current between these two will be 1 A, and current supplied to load is again 10 A. So here P=VI holds, as the load is not changing? $\endgroup$ Commented Nov 5, 2023 at 16:36
-
$\begingroup$ This is correct, assuming the line resistance is negligible. If you now assume that there is a small resistance $R$ connecting the transformers, there will be an $I^2R$ loss where $I \approx 1\text{ A}$. The point is the higher the step-up/down ratio, the lower this current, and hence the lower the transmission loss. $\endgroup$– PukCommented Nov 5, 2023 at 16:44