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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

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  • 3
    $\begingroup$ Wud u plz rite like an adult? $\endgroup$ – Alfred Centauri Sep 23 '14 at 11:50
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Circuit Diagram

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 d and from a to + terminal and e to - terminal $P.D \approx 0V$ .
Though between b and c $P.D = 2V$ and between d and e $P.D = 4V$
$$V=\frac{E}{Q}$$ For every $1 C$ of charge passed there must be loss of $1 J$ to cause potential difference of $1V$. 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,$E\approx 0$.

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  • $\begingroup$ Another question.What decides how the energy is lost?When the electrons are almost near b how do they just know there's another resistor ahead? If only the 2v resistor was in the circuit all the energy would be taken by the 2 v resistor and the extra will dissipate as heat. Why can't that happen here? $\endgroup$ – SMcCK Sep 25 '14 at 7:32
  • $\begingroup$ @SmCck you are forgetting that V does not mean energy but rather voltage which is energy/charge. Electrons at b do not know how many resistors are ahead but actually there will be less electrons/time as resistance increase which essentially means that due to less current, which is charge PER TIME, there will be smaller energy dissipated PER TIME. $\endgroup$ – Suchal Sep 25 '14 at 7:46
  • $\begingroup$ So if I increased the voltage ,the energy/time would increase leading to the charge/time to increase.And voltage drops will always balance such that it adds up to the total voltage provided by the battery. $\endgroup$ – SMcCK Sep 25 '14 at 8:12
  • $\begingroup$ How do you propose to increase energy/time? charge/time=current. to increase current you will either need to decrease resistance or to use battery with higher e.m.f. $\endgroup$ – Suchal Sep 25 '14 at 8:30
  • $\begingroup$ total voltage will be equal to voltage of each resistor because: total energy is conserved which means that energy provided by battery = energy used up in resistors that means that energy provided by battery PER UNIT CHARGE = energy used up in resistors PER UNIT CHARGE and energy per unit charge = voltage. does that help? I can't figure out exactly what is your question. Keep on asking until it feels natural. It is a good thing to ask question as long as you have even a slightest confusion. $\endgroup$ – Suchal Sep 25 '14 at 8:33

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