Nice try, except your signs are wrong. Your equation would give R = -4.56K
How do you figure the signs are wrong?Nice try, except your signs are wrong. Your equation would give R = -4.56K
Your idea is identical to mine except for your wrong signs.
It depends on what you are trying to do.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.
I don't have any formal training other than Ohms' Law. I'm more suck it and see or trial and error.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?
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