How can I use circuit analysis to prove RI = RSRP/(RS - RP)?

[whatisgoingon]

Homework Statement

I have to prove that R_I = R_SR_P/(R_S - R_P) by circuit analysis.

The Attempt at a Solution

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So I tried to create a system of equations to do this:

For the first circuit using kirchhoffs rules:
V-I_A(R_S + R_I) = 0 [1]

For the second circuit:

V-I_BR_S - I_PR_P = 0 [2]

V-I_BR_S - (I_A)/2 * R_I = 0 (It's I_A/2 because of kirchhoffs rules regarding the junction therefore it's equivalent to the first circuit's I divided by 2) [3]

I_B = I_P + I_A/2 [4]

First I solved for V for first 3 equations. Then for equation 4, i solved for I_A/2 to make it easier for me, then I substituted it and tried to eliminate as much variables as I can by equating [1] and [2]. then whatever result I get for that, I then equate it to [3]. After doing this, I get
I_B(2R_s + R_I) = I_P(2R_s + R_I -R_P)

Solving this using my head, I know I won't be able to prove the Ri equation. Any tips or other methods to do this?

 

[gneill]

Are there any particular circuit conditions under which this expression is meant to hold? The problem statement doesn't mention anything, but to me these are just two completely independent circuits with variable resistors. Why variable?

 

[whatisgoingon]

I'm not really sure what you mean but this is from a lab and for the series circuit, full scale deflection of the meter is required. The parallel circuit, one-half scale deflection of the meter was required, which is why the parallel resistor was added.I originally used the parallel circuit to try to prove the Ri equation but I ended up with:
R_I=R_SR_P/(R_P - R_S) which is wrong.
So I asked my professor, and this [the attempted solution: using both circuits] is what he hinted me to do and I just can't seem to solve it.

 

[gneill]

Well, that's context that is important! So somehow you have to incorporate the meter conditions into the analysis. What does half deflection versus full deflection imply? (think: current)

EDIT: Or think voltage across the meter. Either way works.

 

[whatisgoingon]

I 'm sort of getting it now. Since the first circuit is in series, lowering the resistance increased the deflection. Using Ohm's law, I =V/R, the meter indicated more current was able to flow through. For the 2nd circuit, using kirchhoffs law, increasing the resistance by a little bit, decreased the deflection by signifcantly since it was in parallel. therefore the current in the meter decreased.

I still can't think of a way to incorporate that into the analysis...Right now my mind is heading towards something about manipulating the current but I don't know how to do it in terms of the equations

 

[gneill]

Start by writing expressions for the current through the meter for both circuits.

 

[rude man]

The only thing making sense to me is:
1. using circuit 1, record R_s1 and i_l.
2.Using circuit 2, adjust R_s2 and R_p for same i_L as for ckt. 1. Record R_s2 and R_p.
Then you can express R_l in terms of R_s1, R_s2 and R_p. R_s2 will be < R_s1.