There is this simple test:
Three identical bulbs are connected in the circuit illustrated in the figure. When switch $S$ is closed:
a] The brightness of $A$ and $B$ remains the same, while $C$ goes out.
b] The brightness of $A$ and $B$ remains the same, while that of $C$ is halved.
c] The brightness of $A$ and $B$ decreases while $C$ goes off.
d] The brightness of $A$ and $B$ increases while $C$ goes off.
For my opinion the answer to this question is D because the switch (which has a resistance of $0\, \Omega$ has a node connected before the third bulb C) that "interrupts" the circuit. But, going into detail, according to Kirchhoff's first law the current should also go on the third bulb as in the first red node it divides into two currents $I_1$ and $I_2$. The current $I_1$ goes for example in the key $S$ and $I_2$ in the third bulb. The key and the third bulb have the same potential difference. I believe that the current $I_2$ passes through the third bulb but the current passing through it is so small that it does not turn on.
I made a point. When an individual is operated on at the heart and puts a by-pass (a bridge), blood will flow on the tube that detects the by-pass and the occluded artery (the third bulb) where blood will flow slowly, over time it will atrophy.
If the circuit were like the one drawn in the picture I would answer the b).
My question is: I have not very clear the rule of a switch in a eletric-circuit.
In fact, I find it difficult to give an answer to the following image.
The battery shown in the figure has negligible internal resistance. After closing the switch S, the bulb $B_1$:
a) it becomes brighter.
b) it becomes less bright.
c) It remains as bright as when the switch was open.
d) You can't say with this wiring diagram.