# Kvl on left side

1. Feb 12, 2010

### nhrock3

i cant understand how they how the member of which i point too

2. Feb 12, 2010

### The Electrician

You haven't designated the two inductors, so I'm going to call the inductor which is common to the two loops L1, and the other inductor L2.

The term you're asking about comes about because the current I2 passes through L2 and induces a voltage in L1. The voltage induced in L1 has the value I2*M.

I think you may have made a mistake in calculating M. It looks like you didn't properly take into account the coefficient of coupling which is .8. Shouldn't M be j40 instead of j50?

3. Feb 12, 2010

### nhrock3

L1 is 0.02
L2 is 0.005
is there still a mistake?

4. Feb 12, 2010

### The Electrician

M = .8 * SQRT(.02*.005) = .8 * .01 = .008

then its impedance would be 40j, not 50j.

5. Feb 12, 2010

### nhrock3

so if k=0.8 than we need to add 0.8 I2
this is the affect of the other coil
??

6. Feb 12, 2010

### nhrock3

they say jwM=j*5000*0.01=50

7. Feb 12, 2010

### The Electrician

Look at the computation you have in your image:

0.8 = K = |M|/SQRT(L1*L2) => |M| = .8*SQRT(L1*L2). (What happened to the .8?)

From that M = .008

You should be able to find K = M/SQRT(L1*L2) in any textbook, or here:

http://en.wikipedia.org/wiki/Inductance

Perhaps "they" have made a mistake; it does happen.