- #1
kmarinas86
- 979
- 1
Let's say the current in an inductor goes up like this:
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.
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.
Last edited: