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I am looking through some old exam problems and I came across one with a solution, I don't quite get.

"In the circuit below what is the voltage drop across R3?"

enter image description here

Their answer is zero, and they come to that conclusion the following way:

$$vb=va\cdot \frac{R5}{R4+R5}=\frac{2}{3}\cdot va$$

$$vc=va\cdot \frac{R6}{R2+R6}=\frac{2}{3}\cdot va$$

$$v_{R3}=vc-vb=0$$

My question is how they can just ignore R3 when making the voltage division equations in the first place. Doesn't voltage get distributed to R3 as well? It almost seems like they know it's gonna be 0 volts and therefore they ignore it immediately.

Can anyone explain to me why this is, and if my intution of how voltage division works is wrong?

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  • \$\begingroup\$ Classic 'gotcha' bridge circuit, it's essentially a Wheatstone bridge. have a google at "nodal analysis examples bridge circuit" and you will see many walk throughs, and hopefully understand why the voltage across R3 is 0. \$\endgroup\$
    – Sorenp
    Dec 10 '19 at 12:38
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    \$\begingroup\$ I took time from my ... extremely... busy day at the office, to find you the best source material @Carl .. youtube.com/watch?v=SOBomZJg8Mc here you go. \$\endgroup\$
    – Sorenp
    Dec 10 '19 at 12:49
  • \$\begingroup\$ I appreciate the generosity, thank you! \$\endgroup\$
    – Carl
    Dec 10 '19 at 12:53
  • \$\begingroup\$ You are correct, the answer starts by using the voltage-divider equations, which require that the current through R3 must be zero or they are not valid. Then, these equations are used to show that the current through R3 is zero. Circular logic. \$\endgroup\$ Dec 10 '19 at 15:36
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    \$\begingroup\$ A generic method of analysis that will work for this circuit, regardless of resistor values, is mesh current analysis. Of course, the values of ALL of the resistors would be needed. \$\endgroup\$ Dec 10 '19 at 16:49
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My answer is how they can just ignore R3 when making the voltage division equations in the first place.

In order to make a judgement on this circuit, various techniques can be used and, it is advantageous to the person solving the problem, to choose the method that delivers an answer in the shortest time. The most sensible first method to try (as a thought experiment) is any method that attempts to compare the voltage at node VB with that at node VC with R3 disconnected.

Clearly, within a couple of seconds, an experienced engineer will see that those node voltages are equal and, the impact of this is that no current can flow even if R3 were a short circuit.

You don't even need to form the equations; it's enough to see that the ratio of R4 to R5 is the same as R2 to R6.

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  • \$\begingroup\$ Okay, so my intial thought should be to make R5 an open circuit, and then continue on from there? If this is the case, does this method work every time, for example if instead of a current source in the circuit, it was a voltage source? \$\endgroup\$
    – Carl
    Dec 10 '19 at 12:20
  • \$\begingroup\$ It makes no difference about the power source. It becomes a voltage at node VA and node VA is shared by R4 and R2. \$\endgroup\$
    – Andy aka
    Dec 10 '19 at 12:25
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Look at this from this perspective. Both the branches (R4, R5) and (R2, R6) have the same voltage across them which is va.

Forget about the R3 resistor for the time being. Now, determine the voltage drop (va - vb). Clearly, the voltage vb is defined as per the equation. Similarly, the voltage drop across R2 can be calculated as per the above equation. This means across the resistor R3 there is no voltage drop and so no current flows across it. We can ignore R3 because we know the voltage drop across the entire branch.

This is basically the Wheatstone circuit analysis.

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  • \$\begingroup\$ No, you can't just say "forget about the R3 resistor". The only way you can ignore this resistor is if you are already assuming that the solution will be that the current through R3 is in fact zero, which what you are trying to prove. \$\endgroup\$ Dec 10 '19 at 15:38

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