So, I've encountered a problem that doesn't look that hard, but I simply can't solve it. I have a simple non-ideal transformer circuit and I want to compute V_0. What I was able to find so far are the equations for each loop.

For t >= 0:

Loop 1:

i_1 + 2*(i_1)' + (i_2)' = 6

Loop 2:

3*(i_2)' + 2*i_2 + (i_1)' = 0

Transient analysis for non-ideal transformer

  • \$\begingroup\$ Can you be very formulaically (or numerically) clear about what the voltage source actually is? \$\endgroup\$
    – Andy aka
    Dec 12, 2021 at 15:36
  • \$\begingroup\$ u(t) is a unit step function, which for t<0 it is 0 and for t>=0 its value is 1. So the voltage source is a step function which for t>=0, its value is 6V DC \$\endgroup\$
    – Lucas
    Dec 12, 2021 at 15:38
  • \$\begingroup\$ Is it this: t>=0 its value is 1 or is it this: t>=0, its value is 6V DC \$\endgroup\$
    – Andy aka
    Dec 12, 2021 at 15:47
  • \$\begingroup\$ t>=0, its value is 6V DC \$\endgroup\$
    – Lucas
    Dec 12, 2021 at 15:49

1 Answer 1


Here's a hint on how to find \$I_1\$

enter image description here

Does that help you get started? \$I_2\$ will be the current flowing through the 1.3333 Ω resistor, divided by the inverse of the turns ratio (1.5).

  • \$\begingroup\$ thanks for the answer, I'll see what I can do here. That looks a bit more complicated than I thought, but I'll look into it. \$\endgroup\$
    – Lucas
    Dec 12, 2021 at 16:33
  • \$\begingroup\$ It's my way of doing it - draw and manipulate - it also is less prone to silly math errors. In all my years I've never used the way they trained me at school because they usually suck. It is also a way of explaining a deeper understanding of transformer coupling. \$\endgroup\$
    – Andy aka
    Dec 12, 2021 at 16:40

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