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According to my textbook the following linear circuit

schematic

simulate this circuit – Schematic created using CircuitLab

Follows these KVL equations in phasor forms

\$Vs=(R_2+L_3jw+L_1jw)I_1-jwMI_2\$ for the first loop

\$0= (jwL_2+R_3I+jwL_4)I_2-jwMI_1\$ for the second loop

How would this formula changes when the transformer is ideal?

According to ideal transformer definition \$L_1\$ and \$L_2\$ are infinite. How can this be interpreted in the above equations?

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  • \$\begingroup\$ What do you mean by go to infinity? What goes to infinity? The voltage? The current? The phasor diagram? \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 9, 2017 at 22:08
  • \$\begingroup\$ @laptop2d Inductance goes to infinity \$\endgroup\$
    – user138643
    Commented Feb 9, 2017 at 22:23
  • \$\begingroup\$ Are you talking about the impedance of the inductor and the circuit being open? \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 9, 2017 at 23:53

2 Answers 2

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The easiest: An ideal transformer shows R3 and L4 to the other side as a resistor and an iductor in series, but you must multiply the resistance and the inductance by the square of the winding ratio. You have effectively only one current loop to calculate. The current I2 is I1/the winding ratio.

If there exists a source on the both sides, you must replace in your equations the mutual inductance voltage terms by an unknown voltages Va and Vb. You get more equations by stating the ideality Va/Vb=Na/Nb and Ia/Ib=Nb/Na. Ia means your I1 and Ib means your I2.

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If the impedance of the inductor "goes to infinity" then the circuit is open and no current can flow. If the impedance of a resistor goes to infinity then no current flows.

So yes, the equations would be:

$$ Vs=(R_2+L_3jw+\infty jw)I_1 -jwMI_2 $$ and on the other side: $$0= (\infty+R_3I+jwL_4)I_2-jwMI_1$$

since the other components are very small compared to the infinite indcutors:

$$ Vs=(L_1 )I_1 -jwMI_2 $$ $$ (L_2)I_2=jwMI_1$$

$$ Vs = (L_1 )I_1 - jwM\frac{jw M I_1}{L_2}$$

$$ Vs = (L_1 - jwM^2\frac{jw }{L_2})I_1$$

$$ Vs = (\infty -0)I_1$$ and by taking limits all over the place you can deduce that the current would be zero as L1 approaches infinity.

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  • \$\begingroup\$ My question is about ideal transformer where L1 and L2, the inductances of transformer, goe to infinity in the picture that is shown in my question. \$\endgroup\$
    – user138643
    Commented Feb 10, 2017 at 0:07
  • \$\begingroup\$ That is the equations from the picture in the quesiton \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 10, 2017 at 0:15
  • \$\begingroup\$ If the inductance goes to infinity there is no current flow, it doesn't matter what kind of transformer or inductor it is. The impedance of the circuit also goes to infinity \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 10, 2017 at 0:16
  • \$\begingroup\$ @laptop You should learn that an ideal transformer is a goal - wanted, but never reached. In ideal transformer the mutual induction cancels your so called infinite impedance. The currents are thus tightly coupled by the winding ratio. \$\endgroup\$
    – user136077
    Commented Feb 10, 2017 at 12:12
  • \$\begingroup\$ I know it doesn't make any sense, but that's what the OP wanted \$\endgroup\$
    – Voltage Spike ♦
    Commented Feb 11, 2017 at 8:08

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