# Finding an unknown resistance

#### RBR1317

Joined Nov 13, 2010
522
I think the TS had the correct intuition in the beginning to solve this with nodal analysis. If one sets up the node equation correctly, it can be solved for R in about 30 seconds on a calculator.

#### MrChips

Joined Oct 2, 2009
20,295
I think the TS had the correct intuition in the beginning to solve this with nodal analysis. If one sets up the node equation correctly, it can be solved for R in about 30 seconds on a calculator.
View attachment 196919
Nice try, except your signs are wrong. Your equation would give R = -4.56K

#### WBahn

Joined Mar 31, 2012
25,277
Nice try, except your signs are wrong. Your equation would give R = -4.56K

How do you figure the signs are wrong?

(7 V - 12 V) / (6 kΩ) is the current flowing out of the node through the top resistor.
(7 V) / (10 kΩ) is the current flowing out of the node through the right resistor.
(7 VV) / (6 kΩ) is the current flowing out of the node through the unknown resistor.

Should not the currents flowing out of the node sum to zero?

#### MrChips

Joined Oct 2, 2009
20,295
Oops, I did not notice the 12V and 7V reversed in the first term. You are correct.

Again, it is simply applying KCL.

#### Phil-S

Joined Dec 4, 2015
140
Doesn't it get complicated or am I missing something?
A simple voltage divider with three of the four terms known?
The only complication if you can call it that, is one resistor is two in parallel.

#### WBahn

Joined Mar 31, 2012
25,277
Doesn't it get complicated or am I missing something?
A simple voltage divider with three of the four terms known?
The only complication if you can call it that, is one resistor is two in parallel.
It depends on what you are trying to do.

For the problem as stated (and given the reasonable assumption regarding which node is being driven and which is being monitored), finding the unknown resistance is very straight forward. You can do it formally by applying KCL explicitly, or you can apply it in an ad hoc manner quite effectively.

The voltage across the all three resistors is known and so the current in the two known resistors can be determined immediately. Whatever current flows through the top resistor that doesn't flow through the bottom known resistor must flow through the unknown resistor. Hence we know both the voltage across the unknown resistor and the current through it and can therefore solve for the resistance.

But you can ask several more general and interesting questions and those become more complicated, such as instead of having a known output voltage, what value does the unknown resistor need to be in order to produce an arbitrary output voltage and what are the limits on what that voltage can be?

Or consider the case where the unknown resistor is the load and can very from a very high resistance down to some minimum resistance Rmin. What is the fractional change in the output voltage of the divider over the range of load resistance and what relationship do the two known resistors have to have to Rmin in order to keep that fractional chance below some specified value?

#### Phil-S

Joined Dec 4, 2015
140
It depends on what you are trying to do.

For the problem as stated (and given the reasonable assumption regarding which node is being driven and which is being monitored), finding the unknown resistance is very straight forward. You can do it formally by applying KCL explicitly, or you can apply it in an ad hoc manner quite effectively.

The voltage across the all three resistors is known and so the current in the two known resistors can be determined immediately. Whatever current flows through the top resistor that doesn't flow through the bottom known resistor must flow through the unknown resistor. Hence we know both the voltage across the unknown resistor and the current through it and can therefore solve for the resistance.

But you can ask several more general and interesting questions and those become more complicated, such as instead of having a known output voltage, what value does the unknown resistor need to be in order to produce an arbitrary output voltage and what are the limits on what that voltage can be?

Or consider the case where the unknown resistor is the load and can very from a very high resistance down to some minimum resistance Rmin. What is the fractional change in the output voltage of the divider over the range of load resistance and what relationship do the two known resistors have to have to Rmin in order to keep that fractional chance below some specified value?
I don't have any formal training other than Ohms' Law. I'm more suck it and see or trial and error.
I did look at your detailed and crafted set of equations and compared it with the the original diagram.
I don't do SPICE analysis, but I do get a few resistors out and see what works.
I my experience, theory, apart from Ohms' Law, rarely matches actual performance.
I don't do nodes and KCL means Potassium Chloride, but what the Hell, we all get there eventually.
I think you might admit that the apparently simple question ballooned somewhat, but that makes it all the more interesting.