Hey guys,

Struggling a bit here as I can't seem to find anything online (including in this forum) regarding placing one resistor on one half of a parallel circuit.

The source voltage is 12 volts DC, the first voltmeter (V1) is connected in series with the source, measuring 12 volts. After the first voltmeter (V1) the source lead is split into two paths, forming a parallel circuit. Along first of the parallel leads is the second voltmeter (V2); it is the only component along this parallel lead. Along the second of the parallel leads is a 10 ohm resistor (R1) before and in series with the third voltmeter (V3). The parallel leads then rejoin before the fourth voltmeter (V4) and return to the source's negative.

What would the voltage measurements at voltmeter 2 (V2), voltmeter 3 (V3) and voltmeter 4 (V4) be and why? I appreciate any feedback as I don't have any components I can test this with. Also, if this question is already answered please direct me to that resource.

 

They would probably all be zero. The measurement V1 has dropped all the source voltage.

 

Here's the corrected diagram...

 

hi Ald,
Your diagram must be wrong, everything is shorted out!

E

 

So then I guess there has to be resistance on both wires, otherwise it's a short circuit?

 

hi,
Where was the original drawing posted/printed, any link?
E

 

Lol I made it in Gimp XD... I found the answer I was looking for though: there needs to be resistance on both parallel lines to avoid a short circuit. Also, the current divides into two when a lead is split into two, not the voltage. It's clear I was short-circuiting... Thanks for your replies though .

 

If this is really a homework problem, I would love to see the actual assignment. Both of your diagrams are nonsense. The first shows loops with the voltages not adding up to zero (Kirchoff’s voltage law.). The second shows all the voltmeters shorted out as well as the entire circuit being a short.

Hopefully, neither of these was given to you by the teacher.