Hi Electrician,
I have got what you mean for h21 however I still can't figure out how to get h12. As refer to the previous attached file:
h12 = ((40x40)/110)(1/60+(40||70))
Be careful with your expressions. This won't evaluate properly without another set of parentheses. It should be:
h12 =
((40x40)/110)(1/(60+(40||70)))
As I know when I1 is short circuit the 30 Ohm resistor at I1 will be ignored in the circuit calculation and hence I have only 40 Ohm, the middle 30 Ohm, rigth hand side 40 Ohm and 60 Ohm resistor. But how to get equation for h12?
Please help to advice again.....
Thanks a lot,
Best Regards,
Micheal
You say "As I know when I1 is short circuit the 30 Ohm resistor at I1 will be ignored". Again, be careful. When calculating h12, I1 is
open circuited; if it were short circuited, you couldn't ignore the 30 ohm resistor.
Starting from the right (I2), the middle 30 ohm is in series with the left 40 ohm; that series combination is in parallel with the right 40 ohm. That parallel combination forms a voltage divider with the 60 ohm resistor and you can calculate the voltage appearing at the top of the right 40 ohm resistor.
Then the middle 30 ohm and the left 40 ohm form another voltage divider which further divides the voltage which you calculated at the top of the right 40 ohm resistor.
Apply the second voltage divider to the output of the first voltage divider and you will get h12 = .17
If you want to get the expression from the image in your first post, remember that the formula for two resistors in parallel is "product over the sum" or (R1*R2)/(R1+R2).
The first voltage divider expression is:
D1 = (40||70)/(60 + (40||70))
But (40||70) can be written as (40*70)/(40+70) so the D1 expression can be written:
D1 = (40||70)/1 * 1/(60 + (40||70)) = (40*70)/(40+70) * 1/(60 + (40||70))
The second voltage divider is D2 = 40/(30+40) = 40/70
The product of the two can be written:
Code:
D1*D2 = 40/70 * (40*70)/(40+70) * 1/(60 + (40||70))
{ T1 } { T2 } { T3 }
where I've labeled three terms.
The 70 in the denominator of T1 can cancel the 70 in the numerator of T2, then combine what's left of T1 and T2 to get:
D1*D2 = (40*40)/(40+70) * 1/(60 + (40||70)) = (40*40)/(110) * 1/(60 + (40||70))
It's just a bunch of algebra where they've chosen to combine and simplify in a way that you or I might not have chosen!