J
Jon Slaughter
- Jan 1, 1970
- 0
Your right... lets figure it out then, shall we(as if your going to
following along anyways)?
Since the point was to find a law relating the voltages we have
V1 = L1*I1' + M*I2'
V2 = L2*I2' + M*I1'
Now I1 = I2 + I3 where I3 is the current into the tap: Z2 = 2nd inductor
impedence, Z3 = load impedence. The current divder law gives
I2 = Z3/(Z2 + Z3)*I1 = Z*I1 (Z is unitless)
I2' = Z*I1' + Z'*I1
or
V1 = (L1 + M*Z)*I1' + M*Z'*I1
V2 = (L2*Z + M)*I1' + L2*Z'*I1
Canceling the I1's gives
L2*V1 - M*V2 = [L2*(L1 + M*Z) - M*(L2 + M)]*I1'
= [L1*L2 + M*Z*L2 - M*L2 - M^2]*I1'
With me so far?
We know that M = k*sqrt(L1*L2)
so
L2*V1 - M*V2 = [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'
Now substituting V1 = V - V2 we have
L2*V - L2*V2 - M*V2 = L2*V - (L2 + M)*V2 = ...
Solving for (L2 + M)*V2 gives
(L2 + M)*V2 = L2*V - [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'
Solving for V2 gives
V2 = L2/(L2 + M)*V - [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'/(L2 + M)
V2 = Tau*V - Gamma*I1'
Note if Gamma is 0 or practically 0 then We have a voltage divider effect.
I.e., V2 ~= Tau*V. Tau depending on L1 and L2 and there ratio sets tau(it is
not linear necessarily but we'll investigate that later).
The main thing is to determine Gamma:
There are 4 cases
(Ideal transformer => k = 1, High load => Z = 0)
For an ideal transformer k = 1 so we have
Gamma = M*L2/(L2 + M)
M = sqrt(L1*L2)
or
Gamma = 1/(1/L2 + 1/sqrt(L1*L2))
Note if L2 or L1 is small then Gamma ~= 0. Only for large inductance do we
get a gamma that is large. (note this doesn't necessarily mean the voltage
divider effect out of play as we will see shortly)
Even for 1H inductances we get Gamma = 1/2
But then
V2 = Tau*V - I1/2
But unless Tau*V is compariable to the current being drawn it doesn't have
much of an effect. Take V = 110V and tau = 0.25 and drawing 3A. Which is
approximately 5% off the ideal case.
For your run of the mill transformers obviously L1 and L2 are much much
smaller and obviously the effect will be insiginificant. (although for very
low voltages and high current's the effect is much stronger)
(Ideal transformer => k = 1, Low load => Z = 1)
Obviously with Z = 1 we at most double the value of I1.
For the Ideal transformer and practical configurations, while theoretically
you are correct, for all practical purposes the mutual inductance has no
effect.
(Completely Non-ideal transformer => k = 0, High load => Z = 0)
In this case we have and additional factor of L1*L2 to boost I1. But again,
for all practical cases it is very small. (M = 0 in this case and the other
term gets knocked out)
(Completely Non-ideal transformer => k = 0, Low load => Z = 1)
Again, this just doubles the effect which is still quite small.
--------------------
Conclusion
While you are theoretically correct about the mutual inductance, it just
doesn't have any effect on the voltage dividing effect in practical terms(at
least in all but the most extreme cases).
So we are left with simply only with V2 = Tau*V
Tau is the voltage dividing effect. It is set by choosing the ratio of L1
and L2.
Tau = L2/(L2 + M)
If L2 = 0 then Tau = 0 and this is exactly what we expect
If M = 0 ==> Either no mutual inductance or L1 = 0 then Tau = 1 and this is
exactly what we expect.
The curve is nonlinear
BUT NOTE
What does L2/(L2 + M) look like??
How bout the resistive divider law!!!!!!!!!
Genius huh?
