# Why does the power remain constant during the stepping up and stepping down of current in transformers?

I understand quite a bit of transformers, their structure and other concepts and formulas related to mutual induction. However I'm not able to explain why the electrical power has to be constant during stepping up and down of voltage. This will also help me explain why it is beneficial to transport current at high voltages. Please try to mathematically explain to me why power has to be constant. Or put in other words , How can you prove Current is inversely proportional to voltage where Ohms law states otherwise.

What do you mean by "during stepping up and down"? "Stepping up/down" isn't a transient state in the operation of a transformer. A step-up transformer continuously produces a voltage that is higher than the input voltage: it is always stepping up voltage as long as the transformer is in operation.

If the input and output powers were different, how would you account for the power difference? Where would the power difference go/come from?

An ideal transformer is an idealized pair of inductors that are perfectly coupled, have infinite inductances and no losses. Under these conditions, it follows directly from the governing terminal (I-V) relations of the mutual inductors that the input power is equal to the output power.

In real transformers, the output power is less than the input power, because of a number of loss mechanisms:

1. Losses in winding series resistances
2. Hysteresis (magnetization) losses in the transformer core
3. Eddy current losses in the transformer core

Update:

The equations for perfectly coupled inductors are: $$v_1 = L_1\dot i_1+\sqrt{L_1L_2}\dot i_2$$ $$v_2 = \sqrt{L_1L_2}\dot i_1 + L_2\dot i_2.$$ It follows immediately that $$v_2 = \sqrt{\frac{L_2}{L_1}}v_1$$. Rewriting the first equation, $$\dot i_2 = -\sqrt{\frac{L_1}{L_2}}\dot i_1 + \frac{1}{\sqrt{L_1L_2}}v_1$$ Now with the assumption of very large inductances, the second term is negligible, and $$\dot i_2 = -\sqrt{\frac{L_1}{L_2}}\dot i_1.$$ When the transformer is "off", $$i_1=i_2=0$$, so the integration constant upon integrating this equation is zero: $$i_2 = -\sqrt{\frac{L_1}{L_2}} i_1$$ $$v_1i_1+v_2i_2=0.$$

• I know that I'll have to do work to change power but why is that so.? Please try to mathematically explain to me why power has to be constant. Or put in other words , How can you prove Current is inversely proportional to voltage where Ohms law states otherwise. Jun 23 '20 at 7:22
• See the update. Ohm's law applies to resistors, not inductors.
– Puk
Jun 23 '20 at 7:44

However I'm not able to explain why the electrical power has to be constant during stepping up and down of voltage.

For an ideal (lossless) transformer energy must be conserved, i.e., power into the primary = power out of the secondary, or P=VI= constant. A transformer doesn't generate electrical energy.

Or put in other words , How can you prove Current is inversely proportional to voltage where Ohms law states otherwise.

There is no conflict with Ohms law. Ohm's law provides the relationship between voltage, current and resistance.

$$P=VI$$

For a resistor

$$P=I^{2}R$$

Therefore, for a resistor

$$VI=I^{2}R$$

which returns Ohm's law

$$V=IR$$

Hope this helps.

• Precise. Law of conservation of energy was the word I was looking for And I was too silly not to have thought of power as energy/time. BTW, I know the derivation of the equation of E. power= V . I. I was just confused that when power is constant and voltage is changing, current changes inversely to the voltage which I thought was a contradiction to Ohms law (but again a transformer's inductor is not a Ohmic conductor. Jun 24 '20 at 4:46

P=E/t .if t is 1 second ( unit time) then p =e (energy) . in a transformer what happens is through a moving voltage, a moving current is passed through the wire so that there is a magnetic field created( biot-savart law{B=miu d (Ix )/4π r^2 ) which results a current (to be precise )due to a back EMF( according to Faraday and lenz law in electromagnetic induction) is created in the other given wire. This is how the other wire also generates current(for further information study electromagnetism deeper). if we ignore all losses of energy in a transformer (not realistic though),then the P is constant due to energy conservation law. But in reality due to Eddy currents joule heating and other things the converted electric energy in the secondary coil is lesser than in the primary coil. Hoping you got the answer for why p is constant. W=VQ. (Q is the charge that flows through V voltage difference in a conducter) I=Q/t So, W = VIt W/t =P P=VI Therefore when p is constant v is inversely proportional to I. ..hope you understand how the equation came into existence which thereby explains this.. I am merely a student learning physics AND still learning, so my answer rather precise to say opinion might want improvement or to be corrected but through pleasure I made the answer..hope it helped though..(also new to the app so might make mistakes while using it.. please correct me if any)