Following on from Mark's comments above, Alan's method is akin to the standard approach to determine the input resistance of an unknown 'black box'. Let's do an analysis for the general case.
Let: the source voltage = Vs; source internal resistance = Rs; input resistance of our 'black box' = Rin.
Prior to introducing the variable resistance between Rs and Rin, let the voltage appearing across Rin be Vin.
We have:
Vs / Vin = (Rs + Rin) / Rin,
which can be re-arranged as:
Vin / Vs = Rin / (Rs + Rin),
so:
Vin = Vs[Rin / (Rs + Rin)] . . . . . eq'n. 1.
We now introduce a variable resistance R between Rs and Rin and adjust its value such that the value of Vin is reduced to Vin/2. Let the value of that variable resistance that produces that figure of Vin/2 be R.
Now we have:
Vs / (Vin/2) = (Rs + Rin + R) / Rin,
which can be re-arranged as:
Vin / (2.Vs) = Rin / (Rs + Rin + R),
so:
Vin = 2.Vs[Rin / (Rs + Rin + R)] . . . . . eq'n. 2.
Since eq'n. 1 and eq'n. 2 are both equal to Vin, we can now write:
Vs[Rin / (Rs + Rin)] = 2.Vs[Rin / (Rs + Rin + R)].
This reduces to:
Rin + Rs = R . . . . eq'n. 3
Now for Rin to be approximately equal to R (our introduced variable resistance), we require Rs << Rin.
This is the usual case where we have a 'black box' with a high input resistance and is connected to a source of voltage whose internal resistance is very much lower. However, if that condition is not satisfied, then Rin = R - Rs.
Hence, in Alan's test arrangement, what needs to be done is to choose a value of R (his R-Box) so that his initial Vin (= 0.9 v) drops to 0.45 v. And if Rs does indeed equal 5 ohms, and Rin also equals 8 ohms, eq'n. 3 predicts that the value of R will be
8 + 5 = 13 ohms.
I sincerely hope all of that (and my preceding post) was not to difficult to follow: simply some elementary electrical theory and a bit of basic algebra.
Al.
Let: the source voltage = Vs; source internal resistance = Rs; input resistance of our 'black box' = Rin.
Prior to introducing the variable resistance between Rs and Rin, let the voltage appearing across Rin be Vin.
We have:
Vs / Vin = (Rs + Rin) / Rin,
which can be re-arranged as:
Vin / Vs = Rin / (Rs + Rin),
so:
Vin = Vs[Rin / (Rs + Rin)] . . . . . eq'n. 1.
We now introduce a variable resistance R between Rs and Rin and adjust its value such that the value of Vin is reduced to Vin/2. Let the value of that variable resistance that produces that figure of Vin/2 be R.
Now we have:
Vs / (Vin/2) = (Rs + Rin + R) / Rin,
which can be re-arranged as:
Vin / (2.Vs) = Rin / (Rs + Rin + R),
so:
Vin = 2.Vs[Rin / (Rs + Rin + R)] . . . . . eq'n. 2.
Since eq'n. 1 and eq'n. 2 are both equal to Vin, we can now write:
Vs[Rin / (Rs + Rin)] = 2.Vs[Rin / (Rs + Rin + R)].
This reduces to:
Rin + Rs = R . . . . eq'n. 3
Now for Rin to be approximately equal to R (our introduced variable resistance), we require Rs << Rin.
This is the usual case where we have a 'black box' with a high input resistance and is connected to a source of voltage whose internal resistance is very much lower. However, if that condition is not satisfied, then Rin = R - Rs.
Hence, in Alan's test arrangement, what needs to be done is to choose a value of R (his R-Box) so that his initial Vin (= 0.9 v) drops to 0.45 v. And if Rs does indeed equal 5 ohms, and Rin also equals 8 ohms, eq'n. 3 predicts that the value of R will be
8 + 5 = 13 ohms.
I sincerely hope all of that (and my preceding post) was not to difficult to follow: simply some elementary electrical theory and a bit of basic algebra.
Al.
