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Micheal

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Hi All,

Please refer the attached file, I rather confuse on how to get the h paramater for h12 and h21. I know the formula of it whereby h12 = V1/V2 ( when I1 = 0 ) and h21 = I2/I1 ( V2=0 ). . The reference I get here show the solution but somehow I can't understand how it work?

For h21 = - (40x40) /100x94 = -0.17 ( this is shows in the reference which I didn't attach it )

Please kindly help to explain how to get those value.

Thanks ....

Regards,
Micheal
 

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Hi All,

Please help to explain when I=0 or V=0 what theorem analysis should I apply to calculate h12 and h21?

Confuse and can't figure out.

Appreciate a lot for the kind explaination.

Thanks & Best Regards,
Micheal
 
Hi All,

Please refer the attached file, I rather confuse on how to get the h paramater for h12 and h21. I know the formula of it whereby h12 = V1/V2 ( when I1 = 0 ) and h21 = I2/I1 ( V2=0 ). . The reference I get here show the solution but somehow I can't understand how it work?

For h21 = - (40x40) /100x94 = -0.17 ( this is shows in the reference which I didn't attach it )

Please kindly help to explain how to get those value.

Thanks ....

Regards,
Micheal

You need to remember what the definitions of the four h parameters are.

h21 is the short-circuit forward current transfer ratio. That means, if you inject a current at I1, and measure the current through the 60 ohm resistor when the output is shorted to ground (that current will be I2), then h21 = I2/I1, taking the sign into account. Remember that current into the I2 node is considered positive, so current out of the node will be negative.

You can start at the output and calculate the equivalent resistance of the 60 and 40 ohm resistors at the output in parallel (the I2 node shorted to ground). Then add the middle 30 ohm resistor in series with that combination; now you can use the current divider rule to get the current in the middle 30 ohm resistor. Then apply the current divider rule again to get the current in the 60 ohm resistor.

The definition of h12 is the open-circuit reverse voltage transfer ratio. In this case, apply a voltage at I2 and calculate the resulting voltage at I1, with I1 open-circuited.
 
The definition of h12 is the open-circuit reverse voltage transfer ratio. In this case, apply a voltage at I2 and calculate the resulting voltage at I1, with I1 open-circuited.

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))

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
 
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!
 
Last edited:
Hi Electrician,

Thanks for the great explaination.

Btw, if the question without the 60 Ohm resistor in the circuit as attached previously, to find h21 whereby V2 = 0. Can I say that there will be additional loop of cable which is shorted beside the right hand side of 40 Ohm resistor at V2 ?

Please advice,

Thanks,
Micheal
 
Yes. Then the output current I2 will be the current in that shorted loop of cable. The right hand 40 ohm resistor will no longer be involved in the expression for h21.
 
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