3
\$\begingroup\$

Ohm's law states V=I*R.

That means when we increase voltage we must also increase the current(I.)

But transformer increases the current while decreasing the voltage or decreasing the current while increasing the voltage.

How does this happen?

\$\endgroup\$
  • 2
    \$\begingroup\$ Because at best what you can get is Pin = Pout (Vin x Iin = Vout x Iout ) 100% efficiency. \$\endgroup\$ – G36 Jun 2 at 10:56
  • 18
    \$\begingroup\$ Ohms law states V=I*R Sure, but that applies to resistors and not transformers. \$\endgroup\$ – Bimpelrekkie Jun 2 at 11:01
  • 1
    \$\begingroup\$ Two words: Lenz law. \$\endgroup\$ – winny Jun 2 at 16:57
  • 1
    \$\begingroup\$ @Bimpelrekkie OL can be applied to everything, it's just useless for non-ohmic situations. In steady state (constant DC current) OL is completely valid for a xformer electronics.stackexchange.com/questions/339055/… \$\endgroup\$ – vaxquis Jun 3 at 10:14
  • 1
    \$\begingroup\$ @vaxquis constant DC current I do not disagree, however what is the functionality of a transformer at "constant DC current"? The behavior of a transformer at "constant DC current" does not bear any direct relation to its behavior at AC currents. \$\endgroup\$ – Bimpelrekkie Jun 3 at 10:35
23
\$\begingroup\$

Ohmls Law states V = IR. That means when we increase voltage we must also increase the current (I).

That is true when feeding a resistor.

But transformer increases the current while decreasing the voltage or decreasing the current while increasing the voltage.

A transformer is not a resistor so you can't use Ohm's law on it.

How does it happen?

A transformer is an electrical gearbox.

        | In                      | Out
--------+-------------------------+-------------------------
Gearbox | High speed, low torque. | Low speed, high torque.
Trafo   | High V, low I           | Low V, high I

It's important to realise that (ignoring losses) power in = power out. From the Joule-Lenz Law we know that P = VI so if V is reduced I must increase inversely.

\$\endgroup\$
  • 1
    \$\begingroup\$ nitpick: you can use OL, it's just useless - the relation between V, I and R is still valid, it's just that actual momentary value of R in a coil varies in relation to V & I... same as with diodes, transistors etc. \$\endgroup\$ – vaxquis Jun 3 at 10:13
  • 3
    \$\begingroup\$ Thanks for the feedback. I pitched the answer at the same level as the question. \$\endgroup\$ – Transistor Jun 3 at 10:18
  • \$\begingroup\$ so you are saying ohms law does not work in ac circuits or in transformer based circuit \$\endgroup\$ – Jyotir Jun 3 at 15:14
  • \$\begingroup\$ No, I did not say that. You can use Ohm's Law (note capitals) on AC circuits on resistive or reactive (L or C) elements. A transformer is not in that category although it can be modeled using R, L and C along with an ideal transformer so you generally don't use Ohm's Law on the transformer itself.. \$\endgroup\$ – Transistor Jun 3 at 16:51
  • \$\begingroup\$ Thanks man i am out of dillema now \$\endgroup\$ – Jyotir Jun 5 at 15:55
8
\$\begingroup\$

"when we increase voltage we must also increase the current(I)" while R is constant.

You should look at the transformer from a power perspective: P=I*V

and Power In = Power Out,

Now, if you have 10V in and 1 A then that is 10W, so then power out is 10W

If you have 10 times the number of turns on the output compared to the input side then you will get 100V but at 0.1A ie 100*0.1 is 10W.

If you have 10 times the number of turns on the input compared to the output side then you will get 1V but at 10A ie 1*10 is 10W.

The wire used for each winding has to have sufficient thickness ie thicker for higher current. Any losses have been ignored.

\$\endgroup\$
7
\$\begingroup\$

The "left" side of the transformer (the side the voltage is applied to) obeys Ohm's law (technically a generalised form that describes impedance instead of just resistance). The currents and voltages that don't seem to obey Ohm's law happen on the other side of the transformer, in an electrically isolated circuit. Ohm's law doesn't describe how two circuits relate, but how voltage relates to current in the same circuit.

\$\endgroup\$
3
\$\begingroup\$

The transformer uses the shared-flux of the core as a negative feedback mechanism. The primary and secondary fluxes ALMOST perfectly cancel, with the residual called the "Magnetizing flux".

If the magnetizing-flux becomes too small, then more energy is taken from the primary (the energy source) and the core flux is again adequate to produce what the secondary is requiring.

Similarly, if the primary has 100 turns with current Ip, and secondary has 300 turns, then the secondary can delivery only 1/3 of the current before the flux generated by the secondary has balanced out (cancelled) the primary flux.

Again, the transformer core is the summation-mechanism for a negative-feedback regulatory system.

\$\endgroup\$
0
\$\begingroup\$

You are confusing the "Lossless Transformer's" function with the resistor's function. The resistor's function is to convert the applied voltage and current flow to thermal energy for dissipation. The transformer's function is to convert an applied input voltage and current to another voltage and current with NO DISSIPATIVE LOSSES. For 10 Watts input at the transformer you will have 10 Watts available at the output. Thus you use a differing model to define the transformer than a resistor.

Obviously a 'Lossless Transformer' only exists in our simulations and thought exercises. But for practical purposes it allows us to use a simple set of rules about voltage and current to define the transformers critical behaviors of interest without resorting to the maddening world of Maxwell's Equations and various other high level math functions. That simplification allows us to use the turns ratio to project the voltages and currents. With that said we know a transformer with 100 turns on the primary and 10 turns on the secondary has a turns ratio of 10. So if the transformer has 100 VAC at the input, the lossless transformer will have 10 Volts at the output. Similarly if 1 Amp is being drawn by the the input winding then the output is delivering 10 Amps to a load. 100 Watts of Power at the input is converted to 100 Watts of Power at the output.

In the real world the windings use wire which exhibit resistance. Power is lost in those wire resistances in both the primary winding and the secondary winding. The Brain Trust of Transformer Designers in over 100 years of designing transformers have developed very efficient cores with low resistance wire providing us with off the shelf transformers reaching over 98% efficiency. There Ohm's law is applicable, but most application level users of transformers can ignore the losses. Of course if you are a utility like ConEdison with generators transmitting 10 MegaWatts that 2% at 10 cents per KiloWatt hours adds up real fast and makes for a very excitable bunch of bean counters.

\$\endgroup\$
0
\$\begingroup\$

The Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across THE (same) two points. It’s applicable to all circuits and transformer is not an exception. A mistake that led to contradiction is that (decreasing) current is measured not between the same points, where (increasing) voltage is. Current is measured in primary winding, but voltage is measured across secondary. If we will measure current and voltage at the same side of transformer we’ll find that the Ohm’s law is still in place. Moreover, if we compare \$\frac{V}{I}\$ ratios at different sides of transformer, we’ll find, that transformer not only changes voltages and currents, but apparent resistance (impedance) as well. For example, if ideal transformer decreases voltage by factor of 2 (turn ratio is 2), and secondary winding is loaded by resistor R, then resistance (impedance) at primary side will appear as \$R\cdot2^2\$. So, apparent resistance transformed by the factor of turns ratio squared.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.