Using only intuition, without mathematics, why can't the total resistance in the circuit of the 1st image be less than the total resistance in the 2nd image?
The 2nd circuit gives more path options for a current to flow, but I don't see why this could result in less resistance than the 1st circuit.
Without doing any actual calculations, consider an extreme case.
Let a be a small resistance, say 1 ohm. Let b be a large resistance, say 1M ohm.
In the first diagram, a becomes irrelevant, and we have two big resistors in parallel.
In the second diagram, b becomes irrelevant, and we have two small resistors in series.
If we work it out, the first case gives 500k ohm, the second is 2 ohm.
What if a is lower resistance than b? Assume a = 100 ohms and b = 1000 ohms. Do the calculations.
Now assume a = b = 100.
I think you'll find something interesting.
Your intuition is correct, but for some reason you did not realize that if, in your words, there is "more path option for a current to flow", then that means the resistance is lower. Ohm's law says, "less resistance; more current".
With the switch open, consider the voltages on P and Q.
Case 1 - Vp = Vq.
When the switch closes, no current flows through it. The conditions at A and B terminals are unaltered.
Case 2 - Vp ≠ Vq
When the switch closes, current flows, which extra current comes from the A and B terminals, and more current implies less resistance.
Assuming resistance a >>> b
Case 1
The equivalent resistance A-B would be two 'a' resistors in parallel.
Case 2
The equivalent resistance A1-B1 would be two 'b' resistors in series.
Hence resistance A-B would be greater than resistance A1-B1 except in the case of a = b when both resistance A-B and A1-B1 would be the same.
