I have a circuit with a 6 volt battery and 2 resistors A and B in that order.A has a potential drop of 2 volts and B has a P.D of 4 v. which means every electron moving across the resp. resistors would lose the resp. amt of energy(which is the P.D).
Now i grab a voltmeter :
1) The voltmeter reading across the battery wud be 6v bcoz an electron gains 6v of energy when moving across the battery.
2) Now i connect the voltmeter to any 2 points before resistor A. and yes i take the insulation off. note that there isn't a resistor between the 2 points of the voltmeter's connections.it's just plain wire. since hardly any energy is lost by an electron while travelling in the connecting wires, the electric potential at those two points would be the same. which means my potential DIFFERENCE wud be 0 or close to 0. so will the voltmeter read 0 when connected to 2 points in the connecting wire? plz explain if wrong.
3)And what wud the reading be at two points BETWEEN resistor A and B?. Again, just plain wire
4)And after resistor B? (yes just plain wire)
Plz relate your answer to the energy lost by an electron.I have heard of countless analogies(including something abt a sandwich) but they annoy me. i want to get to the atomic level. And oh yes
PLZ ANSWER
Best Answer
We always consider than wire has negligible resistance and if the wire is 99% copper wire it usually has resistance too small to cause drop in voltage measurable by common multimeter used in laboratory.
Considering that the wire you used was such wire, between
a and b,c and dand froma to + terminalande to - terminal$P.D \approx 0V$ .Though between
b and c$P.D = 2V$ and betweend and e$P.D = 4V$$$V=\frac{E}{Q}$$ For every $1 C$ of
chargepassed there must be loss of $1 J$ to causepotential differenceof $1V$. It means $6.24 \times 10 ^{-18}$ electrons lose $1 J$ of energy(common analogy say by bumping into the ions.) That means that when $P.D \approx 0$,change in energy,$E\approx 0$.