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How does energy remain conserved in a transformer if emf is increasing, or decreasing? Does the current decreases to accomodate? Does Ohm's law still hold here? Although we know, Ohm's law is not universal.

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If you're concerned about Ohm's law, you might be getting a little muddled in your thoughts (don't worry, I only know this because I've had my turn at being muddled before on this point!)

Ohm's law doesn't describe a transformer: Ohm's law may apply to whatever load you connect to the transformer; if so, then Ohm's law combines with the transformer law $V_P\, I_P = V_S\, I_S$ to yield the concept of reflected impedance (as I talk about in the linked answer): a load with impedance $\mathcal{Z}$ connected across the transformer's secondary appears as the load $\left(\frac{N_S}{N_P}\right)^2\,\mathcal{Z}$ at the primary. This generalises to:

$$Z_P \approx \left(Z_{T1}^{-1}+\left(Z_{T2}+\left(\frac{N_S}{N_P}\right)^2\,\mathcal{Z}\right)^{-1}\right)^{-1}$$

in a real transformer, where $Z_{T1}$ and $Z_{T2}$ model the transformer's own ohmic losses and energy storage capacity.

The transformer equation itself can be thought of as either:

  1. A statement of conservation of energy; OR
  2. A statement that the same flux links the primary and secondary and the consequences of applying (a) Ampère's law to a flux loop threading both primary and secondary to get the current ratio from and (b) Faraday's law to flux time variation, to get the voltage ratio.
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Yes you're right. Wikipedia calls this the transformer equation: $$\frac{V_P}{V_S} = \frac{I_S}{I_P}=\frac{N_\text{P}}{N_\text{S}}=\sqrt{\frac{L_P}{L_S}}$$ $P$ and $S$ denote primary and secondary, respectively. $V,I,N$ and $L$ are emf, current, winding number and inductance. Keeping this ratio constant assures that the power $P=V_SI_S=V_PI_P$ is the same on either side.

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  • $\begingroup$ Yes, I figured that out long ago. But what I could figure out since long was that, wouldnt it violate our old pal Ohm's law? $\endgroup$ – Rohit Jul 4 '14 at 1:51
  • $\begingroup$ @Rohit: No, this becomes a voltage to be considered. So if you plug a $10:1$ stepdown transformer into a $120VAC$ outlet, you have $12VAC$ at the output. If you connect that across a $2\Omega$ resistor, you will draw $6 AAC$. That will draw $0.6 AAC$ from the wall. The back emf is then $120VAC$ and all is well. $\endgroup$ – Ross Millikan Aug 2 '14 at 2:38
  • $\begingroup$ @Rohit If you're concerned about how Ohm's law fits in to conservation of energy, please put that in your question. $\endgroup$ – BMS Sep 8 '14 at 6:00

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