- #1

kmarinas86

- 979

- 1

Let's say the current in an inductor goes up like this:

0

6.32

8.64576

9.50163968

9.816603402

9.932510052

And down like this:

10

3.68

1.35424

0.49836032

0.183396598

0.067489948

With an average of 5 amps

The squared current (and therefore instantaneous power) rises as:

0

39.9424

74.74916598

90.28115661

96.36570236

98.65475593

And drops as:

100

13.5424

1.833965978

0.248363009

0.033634312

0.004554893

So the mean square of the amperage is 42.97134159 amps^2 [NOT RMS].

We have two equations:

Transmitted Power=VI

Power Loss=RI^2

Applying these, we would have the following for average:

Transmitted Power=V*5 amps

Power Loss=R*42.97134159 amps^2

If Average transmitted Power > Average power Loss:

V*5 amps > R*42.97134159 amps^2

V/R > 8.594268318 amps

I > 8.594268318 amps

Which is not the case...

*amps*0

6.32

8.64576

9.50163968

9.816603402

9.932510052

And down like this:

*amps*10

3.68

1.35424

0.49836032

0.183396598

0.067489948

With an average of 5 amps

The squared current (and therefore instantaneous power) rises as:

*amps^2*0

39.9424

74.74916598

90.28115661

96.36570236

98.65475593

And drops as:

*amps^2*100

13.5424

1.833965978

0.248363009

0.033634312

0.004554893

So the mean square of the amperage is 42.97134159 amps^2 [NOT RMS].

We have two equations:

Transmitted Power=VI

Power Loss=RI^2

Applying these, we would have the following for average:

Transmitted Power=V*5 amps

Power Loss=R*42.97134159 amps^2

If Average transmitted Power > Average power Loss:

V*5 amps > R*42.97134159 amps^2

V/R > 8.594268318 amps

I > 8.594268318 amps

Which is not the case...

**Please help.**
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