I read on a book that when a circuit has an AC voltage source as the input then it's input resistance is
Ri= ui/ii
where ui is the instantaneous input voltage and ii the instantaneous input current.
Isn't this false?
I read on a book that when a circuit has an AC voltage source as the input then it's input resistance is
Ri= ui/ii
where ui is the instantaneous input voltage and ii the instantaneous input current.
Isn't this false?
It's not a definition that I've seen. It doesn't strike me as a good definition in general because the math would get a little dicey at the zero crossings. Otherwise I guess it would be basically correct for a resistive network, even if resistances changed with time.
Yes. Its wrong. Input resistance of an ideal voltage source is Zero. pls click the following link for more iformation
-------------------- The input resistance implied is not that of the source but that of the circuit supplied by the load-as seen by the source and is independent of the source.
In any case, an ideal voltage source is a convenient fiction useful in circuit analysis but hard to find in the real world (sure with fast enough feedback one can come close).
As for the definition- it simply notes that v(t)=R*i(t) rather than v(t)=L*(di/dt) or i(t)=C*(dv(t)/dt That is correct.
If R is constant over the range of possible v(t), i(t) then be happy, Ohm's Law is valid and life is simple(?) .--
Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer
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