Problem 3Prove
Miller's theorem.
Answer:
Referring
to figures 1 and 2 above, we need to prove that
| Z' = |
Z
1 k |
and Z" = Z × |
k
k 1 |
, where k = |
V2
V1 |
The current through
Z in Fig. 1 is
For Fig. 2 to
be equivalent of Fig. 1, the same current I must flow through Z' and
Z". The direction of I is same as in Fig. 1 through Z' and opposite
through Z".
In case of Z',
set the current through it equal to that in Fig 1:
Then, solve for Z'
and divide top and bottom by V1:
| Z' = Z × |
V1
V1 V2 |
= Z × |
1
1 V2/V1 |
In the case of Z",
I flows in opposite direction, so V2/Z" = -I = -(V1-V2) /
Again, solve for
Z" and divide top and bottom by V1:
| Z" = Z × |
V2
V2 V1 |
= Z × |
V2/V1
V2/V1 1 |
Contributor:
Naveen P N
4-01
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