# Coupled Inductors

Discussion in 'Homework Help' started by stobbz, Jan 9, 2011.

1. ### stobbz Thread Starter New Member

Jan 7, 2011
7
0

I'm just looking for some clarification regarding the second loop of this circuit.

Using mesh current analysis, the answer gives the voltage across the j9 inductor as:

-j9 * (I1 - I2)

I would have thought it was:

-j9 * (I2 - I1)

Is it (I1 - I2) becuase it is a voltage rise as opposed to a voltage drop?

Any help appreciated!

2. ### Cerkit Senior Member

Jan 4, 2009
286
3
The first answer is correct. If you remember for components between meshes the component's voltage drop is the impedance of the component times the current in mesh one minus current in mesh two
The subtraction done in brackets

3. ### stobbz Thread Starter New Member

Jan 7, 2011
7
0
Ah ok, I was thinking too much about the procedure I have been using to solve these circuits as opposed to looking at what's actually going on in the circuit.

Makes sense now, thanks.

4. ### Georacer Moderator

Nov 25, 2009
5,150
1,271
I agree that the first answer is the correct one, but for a different reason:
We choose to measure the voltage of j9 assuming that the high voltage is up and the low down, judging by the direction of the arrow (provided that it symbolizes voltage and not current).

Then the overall voltage will be j9*i1-j9*i2 because i1 is going from + to - and i2 from - to +.