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?"
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?
Best Answer
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.