Why does simplifying resistance networks give the right answer?

In the circuit
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why is it true that if we simplify the parallel combination as a resistor with an effective resistance of $R/2$ and then use series for $R/2$ and $R,$ it satisfies the condition of equivalent resistance (i.e. draws same current through the same P. D?) It is easy enough in this case, why is it true in the general case?

So why does reductive analysis work at all?

The general method would be to consider charge and take ratio $V/i.$

Edit

We proved $R_\text{effective} ~=~R_1+R_2+R_3+\cdots$ for resistors in series. A similar expression exists for resistors in parallel. My question is why is true to do the same when BOTH series and parallel are there in the circuit. It is like I am seeking the proof of associative(or commutative?) law for resistances...

• The question seems rhetorical. If you don't get the "right" answer, then you didn't "simplify" the network. – MaxW Jun 12 '18 at 6:12
• Are asking why Thevenin's theorem works? Are you looking for a proof or something else? – Alfred Centauri Jun 12 '18 at 11:42

No matter what arrangement of resistors you have there will be some initial potential difference, $V$ across that connection and from that, there will flow/return a certain current, $I$ that will be consistent with the laws. Hence we can finally say that the equivalent resistance is the ratio of that potential difference to the current, so: $$R=\frac{V}{I}$$ So you may conclude that by definition it is the equivalent resistance.
• It may help to imagine your resistor network to be in a black (insulating) box with two terminals sticking out. If you apply a pd, $V$, across the terminals and find a current, $I$, then you know that the resistance between the terminals is $R=\frac{V}{I}$, though you don't know whether you have a single resistor or a network inside the box. They are literally indistinguishable. – Philip Wood Jun 12 '18 at 11:59