Hello everybody!

My question is theoretic, not practical, but practical answers might help as well!

I read the textbook of this homepage (http://www.allaboutcircuits.com/vol_1/chpt_5/7.html), and it states, that a shorted component has no voltage in serial circuits (for expamle R2 in this example. Second circuit figure). However, I don't understand this because the jumper wire on R2 makes the serial circuit serial+parallel.

When I see the figure, the current has two ways to achieve the next resistance. It can go through both the jumper wire and the R2 resistor. So why R2 has no resistance? Why doesn't drop any vontage on R2? The jumper wire and R2 constitute the parallel part of this circuit, don't they?

Please help me to understand!
(And yeah, I'm beginner, electricity is my new hobby!)

 

But the shorting jumper and the resistor, because they are in parallel, have to have the same voltage across them. If it is an ideal shorting jumper, than any amount of current can flow through it while producing no voltage across it. But if there is no voltage across it, then there is no voltage across the resistor. If there is no voltage across the resistor, then there is no current through it.

Let's assume that the jumper isn't ideal but, instead, has a resistance of 0.3 ohms. When placed across a 300 ohm resistor, 99.9% of the current will flow through it and only 0.1% will flow through the 300 ohm resistor. To the rest of the circuit, the combination of the two will look like a 0.3 ohm resistor (actually, a tad bit less). As the jumper gets more and more ideal, the combination will look more and more like just a perfect piece of wire and the resistor will have less and less current flowing in it.

 

If your equipment could measure the values, you can expect these results when using one foot of 24 AWG standard wire as the jumper....

 

Very useful answers, thank you very much!