Overwhelming confusion regarding electromagnetic induction in Step Up Transformers

Question

Suppose I've two step up transformers $A$ and $B$, each with same number of coils and same length of wire and same material of input wire. But $A$ has an output wire of higher resistance than $B$.

According to the internet and my book, the induced emf doesn't depend on the resistance of material of the output wire. So as long as the power generator configuration and the number of coils in the output and input wires remain the same, the output induced emf will be equal in both $A$ and $B$. The output power cannot be more than input power since that violates the law of conservation of energy.

In transformer $A$, let the input power be $$P_I=V_{I}I_I$$ and the output power be $$P_O=V_OI_O$$ and since $P_I=P_O$(considering an ideal situation), $$V_O > V_I$$ $$\therefore I_O<I_I$$

Now, in the second transformer $B$ with lower resistance output wire, $P'_I=V'_II'_I=P_I$ and the output voltage $V'_O=V_O$ since induced emf doesn't depend on resistance.

But by Ohm's Law, $$I'_O=\frac{V'_O}{R'_O}>I_O$$ Now considering Lenz Law, if $R'_O$ decreases, $I'_O$ increases so opposing force increases and for the same input power, $V'_O$ decreases. But this means that $I'_O$ decreases now, so resistive force decreases, so $V'_O$ increases again??? What is exactly happening, and how is conservation of energy being maintained? I'm getting overwhelmed, please help me with this confusion.

Answer 1

(a) A point about your terminology... Simple transformers have two coils, a primary coil and a secondary coil (or, if you like, an input and an output coil). Each coil consists of multiple turns. [You have been referring to a single turn as a 'coil'. Not a good idea.]

(b) Since there is a current in the A and B secondary coils they must be connected to loads, which we can take to be resistors. So when you say "But 𝐴 has an output wire of higher resistance than 𝐵" I'm interpreting this to mean that its load resistance is greater.

(c) You have $P_I'=V_I'I_I'=P_I$. But you shouldn't be equating $P_I'$ to $P_I$. For transformer B the load resistance is smaller so $I_O'>I_O$. So more power is taken by the load (since, as you point out, $V_O'=V_O$). Therefore more power is taken by the primary (input) coil, that is $P_I'>P_I$.

(d) I'm not going to try and unravel your last paragraph. I'll explain ... Transformer behaviour can be understood to a good approximation by assuming that

• $\frac{V_O}{V_I}=\frac{\text{number of turns on secondary}}{\text{number of turns on primary}}$

• Power in = Power out, so $V_OI_O=V_II_I$

This is the 'black box' approach that you have been using. The first of these 'bullet points' looks like a simple application of Faraday's law, but to understand fully what's going on in terms of electromagnetic induction requires a different approach. A key consideration has to be that transformers don't work unless the current is continually changing (we'll assume alternating sinusoidally). The $V_I, I_I$ etc. that you have been using in the black box approach are usually taken as rms values (a sort of average). When you apply Lenz's law, you need to apply it to the variations in 'instantaneous' current due to its alternating nature.

But you seem to have in mind an additional change in the instantaneous output current, due to a step-change in the load resistance. As well as causing a back-emf in the secondary coil, a back-emf will also be induced in the primary. What happens just after you've changed the load resistance will indeed be quite complicated (especially since changes in pds and currents due to the alternating supply voltage are going on at the same time). However a steady state will soon be restored, with new values of currents in the bulleted equations. I would suggest – and it is only a suggestion – that you don't try to understand these equations in terms of Lenz's law unless you have specifically been asked to do so.