What is the resistance Rx, in order to have Vo/Vs=1/2

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Hi, my teacher asked us to solve this question, keep In mind that i'm not asking for you to do my homework, I think is just a impossible question, i've tried every possible method for solving it with no success, and I really need to solve it for my pontuation.

First i've tried Mesh Analysis with no success.

Second i've tried Node Analysis with no sucess.

Then i've tried Voltage divider with no sucess also.

Mesh, Node and Voltage Divider Analisys

If Rx value is zero, the Voltage Source is short circuited. I don't know what else to do.

  • \$\begingroup\$ Rx = 0 ohms and a typically inane quest to send a student on. \$\endgroup\$
    – Andy aka
    Feb 2, 2017 at 15:37
  • \$\begingroup\$ Write down Vo as a function of Vs. Then try to find the mathematical solution. What if the solution is not exactly correct but very close. For example not Vo/Vs =1/2 but V0/Vs = 0.9999/2. What would be the value of Rx ? In mathematics, everyone wants the exact solution, in Electronics often a close enough solution is all you need. \$\endgroup\$ Feb 2, 2017 at 15:41
  • \$\begingroup\$ I think in this case it just a bad question. Report it back to you prof.. unless you have copied it incorrectly. \$\endgroup\$
    – Eugene Sh.
    Feb 2, 2017 at 15:44
  • 2
    \$\begingroup\$ You should step back a bit, and approach this from a different angle than what you've done. Analyze the circuit without the 10 ohms. What is Vo? If you then add ANY value resistor to the lower branch, what happens to Vo? \$\endgroup\$ Feb 2, 2017 at 15:54
  • 1
    \$\begingroup\$ 0+ indicating a positive number infinitely close to zero ? \$\endgroup\$
    – Tut
    Feb 2, 2017 at 16:37

4 Answers 4


Since you tend towards the correct answer as Rx tends towards 0 but clearly 0 is not a valid solution would Rx = 1/∞ be considered an acceptable answer?

As has been said, not a good question. The only reason for it that I can think of is to drive home to the student that any additions to a circuit including instrumentation added in order to test or verify a circuit will have an impact on the circuit and the measured values. But there are better ways of doing that.


On the off chance that the question isn't actually about analysis, it is conventional in engineering to take decimal values as indicative of tolerances. So although in mathematics, a = 0.5 b means a is exactly half of b, in engineering you take the tolerance to plus or minus half a unit below the given number of significant figures so a = 0.5 b means a < 0.55 b and a > 0.45 b.

If reading tolerances in this way is the intention of the question, then there are values which answer it.


What is the resistance Rx, in order to have Vo/Vs=1/2

no solution to that question.

conceptually, he is asking you at what value will Rx = Rx // 10R.

that Rx does not exist, short of Rx = 0R.


It is possible to solve this using actual relationships between the circuit's components. Most of the given answers have it right without going through such trouble though. The only possibilities to have Vo/Vs = 1/2 are unsatisfying solutions:

 - A  0 Ohm value for Rx wich leads to a short circuit
 - Removing Ro (the 10-Ohm resistor) which changes the terms of the problem

It is indeed possible to quickly reach such a conclusion either with "common" physical sense or a more rigorous approach.

  1. A first method infers a conclusion directly from the exercise's constraint : Vo/Vs = 1/2 In other words, this voltage relationship means the circuit divides the source's voltage in two halves :

    • A 1st "Vs/2" voltage across the isolated Rx resistor terminals
    • A 2nd "Vs/2" voltage across the two resistors Rx and Ro in parallel

From there, it is easy to conclude the impossibility to solve this, since two different resistive circuits with different values (the single Rx and the paralleled Rx and 10-Ohm) would have an equal voltage at their terminals. As a reminder, the paralleled resistors have an equivalent resistance equal to 1/(1/Rx + 1/Ro), obviously different from Rx (unless Ro has an infinite resistance, ie. is equivalent to an open circuit, or simply said no branch at all)

It is therefore a voltage divider problem, and the only way to divide Vs in two equal halves with the same Rx resistance is as follows :


simulate this circuit – Schematic created using CircuitLab

The solution above is actually equivalent to having an infinite value for Ro, which is like simply removing it. The other possibility would be to not use the same Rx resistor in the two branches. It is then funny (perverse?) your teacher framed the problem through adjusting Rx's value when it is actually the fixed 10-Ohm resistor value which should be changed...?

  1. A second method describes the physical relationship the components have between each other, through Kirchhoff's laws. From this diagram :


simulate this circuit

You can infer the following relationships :

      - (1) Vs = Vx - Vo (mesh rule)
      - (2) Vx/Rx = Vo/Rx + Vo/Ro (sum of currents from Ohm law and junction rule)
      - (3) Vo/Vs = 1/2 (constraint your teacher introduced)

The problem has two parameters : the resistance value of Ro (10 Ohm in this case) and the ratio between Vo and Vs (let's call it "alpha") For the sake of genericity, I replaced the fixed 10 Ohm value by Ro. The answer can thus be generalized for any values. Using the three equations above, and substituting Vx with Vs - Vo you arrive at the following result :

      Rx/Ro = Vs/Vo - 2
      Rx = Ro(Vs/Vo - 2)

Now we have a direct relationship between Ro and Rx so that the coefficient between Vs and Vo is respected. As you can see, the value of "2" you teacher gave for Vs/Vo is the trivial solution, which yields "Rx = 0"

Had your teacher asked the same question for a Vo/Vs ratio of 1/5, then we'd have :

Vs/Vo = 5
Rx = Ro (5 - 2) = 3 x Ro

With our 10 Ohm value for Ro, Rx would require a 30 Ohm resistance.

Finally, here's a plot of Rx against alpha values : Rx values against "alpha" ratio

> It has the form of an invert function
> It yields positive (Rx) values only for alpha ratios between 0 and 1/2 (both of which we know are exluded : the former implies a 0V value for Vo, the latter we proved was a short circuit)

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