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Imagine an ideal transformer interfacing an ideal current source with a ramp function \$i_1(t)=a t\$ connected to the primary winding and a resistor \$R\$ connected to the second winding. We want to know the current in the second loop \$i_2\$.

We could proceed in two ways,

  1. We compute the current in the secondary loop, using ideal transformer's theory we obtain $$i_2 = \frac{N_1}{N_2} i_1 \tag 1 $$.

  2. We could also reason that in an ideal transformer, the magnetic flux in the secondary is the same as in the primary \$\phi_2 = \phi_1 \$, with $$ \phi_1 = B_1 A_1 = \mu n_1 i_1 A_1 \tag 2 $$ with \$n_1 = N_1 / l_1\$. We can then use Ohm's and Faraday's laws to derive \$i_2\$, $$ i_2 = \frac{v_2}{R} = - \frac{\mathcal{E}_2}{R} = - \frac{1}{R} N_2 \frac{d \phi_2}{d t} = \frac{\mu n_1 N_2}{R} \frac{d i_1}{d t} \tag 3 $$ which is clearly different from the expression obtained in equation (1).

Can somebody tell me what is wrong in this reasoning?

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  • \$\begingroup\$ First, since not everyone uses the same terms, please define your \$n_1\$, \$N_1\$, \$l_1\$, etc. Second, please break (3) down into it's constituent steps. I think you're not doing the last step of (3) correctly, but "the last step of (3)" is harder to write about than "(6)". \$\endgroup\$
    – TimWescott
    Dec 8, 2021 at 0:33
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    \$\begingroup\$ Flux through a winding depends on currents in other windings, too. The idea of the transformer is to have as tight as possible coupling between the windings and you do not use it in the calculations. \$\endgroup\$
    – user136077
    Dec 8, 2021 at 0:39
  • \$\begingroup\$ Thanks @timwescott. I didn't know how to number the equations. \$\endgroup\$ Dec 8, 2021 at 8:02
  • \$\begingroup\$ Faraday's law does not talk about the induction of current. \$\endgroup\$
    – Andy aka
    Dec 8, 2021 at 8:11

1 Answer 1

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An ideal transformer has infinite magnetizing inductance. Therefore with finite voltage applied, if the secondary is open circuit, no current flows. You can't apply I =a.t to an ideal transformer.

When the secondary is loaded, current does flow. N1.i1==N2.i2 is true, but the net flux in the core is 0.

Thus your (1) is the correct behaviour; (2) doesn't apply

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    \$\begingroup\$ Thank you very much. Everything makes sense when you realize that the flux inside the ideal transformer is zero. \$\endgroup\$ Dec 8, 2021 at 7:09

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