# A question regarding a transformer

#### AlexK

Joined May 23, 2007
34
Suppose I want to create some sort of an "inductive kick" using a transformer, however instead of disconnecting a battery I prefer using a very fast discharge RL circuit.

Suppose I have a constant current source, I,connected to the primary winding. After a long time a switch disconnects the primary from the source and connects it to a parallel resistor R (thus forming an RL circuit).
So now the current equation is IL=Iexp(-t/tau).
My question is this: If tau is very small (meaning the resistor R is large), so the discharge time is very small, will this create a pulse on the secondary winding (similar to the inductive kick, where the current goes rappidly from constant to zero)?

Thanks.

#### beenthere

Joined Apr 20, 2004
15,819
You will get a kick, but the resistive path will pull current and make the inductive kick smaller. The kick is powered by the collapse in the magnetic field surrounding the transformer winding.

#### AlexK

Joined May 23, 2007
34
What do you mean by "pull current"? Won't the current behave according to the equation I wrote above? And how much smaller will it make the kick?

Suppose I make the following (very) approximate calculation.

flux=B*A; A- cross section area
d(flux)/dt=k*dIL/dt=k*I0*exp(-t/tau)*(-1/tau)

where 'k' is the product of all the constants.
So the e.m.f induced in the secondary winding at the moment t=0 is:

E=N2*d(flux)/dt=N2*k*I0*(-1/tau);

So by choosing tau very small won't I be able to get a high E value according to the above equation?

#### beenthere

Joined Apr 20, 2004
15,819
The resistor in parallel with the winding will make a current path. You get your inductive kick when the induced magnetic field collapses in a shot period and inpuces a much higher voltage in the winding. Allowing a resistor to make a path for current outside of the transformer is going to reduce the inductive kick.