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## Homework Statement

A coil of self inductance [tex]L_{1}[/tex] is coupled to a second coil of self inductance [tex]L_{2}[/tex] and resitance [tex]R[/tex]. The coefficient of mutual inductance is [tex]M[/tex]. The second coil is short circuited whereas the first coil is connected via a switch, to a voltage source [tex]V_{0}cos \omega t [/tex]

## Homework Equations

The switch is closed at time t = 0. Find an expression for the current which then flows in the second coil, as a function of time.

## The Attempt at a Solution

[tex] V_{0}cos \omega t = L \frac {di_{1}}{dt} [/tex]

[tex] M \frac {di_{1}}{dt}= - i_{2}R [/tex]

Transferring into Laplace domain:

[tex] I_{1}(s) = \frac {v_{0}}{L_{1}} \frac {1}{s^{2}+\omega^{2}}[/tex]

[tex] I_{2}(s) = \frac {-Mv_{0}}{L_{1}R} \frac {1}{s^{2}+\omega^{2}} [/tex]

and back to the real world:

[tex] i_{2} = \frac {-Mv_{0}}{L_{1}R\omega} sin \omega t [/tex]

and that just doesn't feel right.

Thanks, in advance.

EDIT: Sorry, lost an s somewhere along the way, got [tex] \frac {-Mv_{0}}{L_{1}R} cos \omega t [/tex] now.

I think that's correct, but wouldn't mind somebody reassurance by someone more experienced.

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