scilearner Posted April 11, 2008 Share Posted April 11, 2008 Hello guys , I got a small question with voltage and charge. If voltage provides potential energy for charges why doesn't the voltage decrease all the time when the charges are moving. I mean I only see voltage change when their are resistors and stuff but my question is shouldn't the voltage change all the time because the charges are losing potential energy as they are moving. Also can you explain potential difference to me. I'm bit confused with that. My electronic undertstanding is not really good and your help would be really appreciated:-):-):-) Link to comment Share on other sites More sharing options...
YT2095 Posted April 11, 2008 Share Posted April 11, 2008 what makes you think the voltage doesn`t decrease? even a length of copper wire has a resistance, all conductors do (except Super conductors of course). the only difference is that Resistor will limit the current more effectively for it`s size than a huge roll of copper wire 2Kms long or something Link to comment Share on other sites More sharing options...
scilearner Posted April 11, 2008 Author Share Posted April 11, 2008 what makes you think the voltage doesn`t decrease?even a length of copper wire has a resistance, all conductors do (except Super conductors of course). the only difference is that Resistor will limit the current more effectively for it`s size than a huge roll of copper wire 2Kms long or something Thank you for the help:-). Well my question is in question I see the volatge always the same. For example in a parallel circuit the resistor recieve the same voltage but when the charges move and you use potential energy how can they have the same voltage. EDIT: I can understand how resistors change the voltage but shouldn't the volatge drop even without resistors. I mean if the charges are moving around and losing potential energy shouldn't the voltage change . I know you are right but I'm confused with this. Sorry for the strange question. Link to comment Share on other sites More sharing options...
swansont Posted April 11, 2008 Share Posted April 11, 2008 It's an idealization to make circuit analysis easier — no potential drop in the wires. In many cases it's a reasonable approximation. Link to comment Share on other sites More sharing options...
Jacques Posted April 11, 2008 Share Posted April 11, 2008 I think that the part you are missing is you have a power source. The voltage would drop to zero fast if there were no current source. For example in a parallel circuit the resistor recieve the same voltage but when the charges move and you use potential energy how can they have the same voltage. The voltage is maintain by the current source. The difference in potential energy use by different resistor depend on the current that goes throug each. The higher the resistance the less current goes throught. PS if the current source is not powerfull enough the voltage will drop. Link to comment Share on other sites More sharing options...
YT2095 Posted April 11, 2008 Share Posted April 11, 2008 as for PD, it would take me about 3 mins with pencil and paper to show you if you were here, but it`s very hard to explain in text! imagine you have a 10 volt power source. and 2 Resistors wired as: +10v---/\/\/\/\/-----(midway test point)----/\/\/\/\/-----0v the resistor on the left is (A) the one on the right is (B) A = 3 Ohms and B=7 Ohms totaling 10 Ohms between them, and using Ohms Law we know that 1 amp is dropped across them (they are using 1 Amp and probably getting warm). P.D is the Difference between any 2 points in that circuit. across resistor A (between the midway and the +10v side)there will be a 3 volt reading if you put a meter across it, across B there will be 7 Volts (again, the Midway and the 0 volt side). across them BOTH will be 10 volts. Link to comment Share on other sites More sharing options...
thedarkshade Posted April 11, 2008 Share Posted April 11, 2008 Also can you explain potential difference to me. Potential difference is the same as voltage. You need to have a potential difference ([math]\Delta\phi[/math]) in order to have the flow of electricity. In electrostatics you might have had the chance to see [math]\phi=k\frac{q}{r}[/math] and [math]U=El[/math] and it's actually [math]\Delta\phi=U[/math] Link to comment Share on other sites More sharing options...
scilearner Posted April 12, 2008 Author Share Posted April 12, 2008 Thanks for everyone who replied:-). I only got a bit of time so I'll rephrase my question. The continued power source makes sense. My question in other words is doesn't the electrical potential energy turn into electrical kinetic energy as the charges are moving. If so why is the voltage constant. Is it because of constant power source. What do resistors really do. Link to comment Share on other sites More sharing options...
imp Posted April 12, 2008 Share Posted April 12, 2008 Thanks for everyone who replied:-). I only got a bit of time so I'll rephrase my question. The continued power source makes sense. My question in other wordsis doesn't the electrical potential energy turn into electrical kinetic energy as the charges are moving. If so why is the voltage constant. Is it because of constant power source. What do resistors really do. Simply remember this: Voltage (potential difference) exists BETWEEN TWO POINTS. It does NOT "turn into kinetic energy". EXISTENCE of voltage CAUSES movement of charge (current). Voltage is only "constant" in the case of a voltage source having unlimited energy- zero internal resistance- an impossibility. imp Link to comment Share on other sites More sharing options...
scilearner Posted April 12, 2008 Author Share Posted April 12, 2008 Thanks everyone for your help now I understand what is going on. All this time I have being thinking about kinetic energy and confused my self. . So basically the only way voltage can drop is due to resistance isn't it. Thanks again!! Link to comment Share on other sites More sharing options...
thedarkshade Posted April 13, 2008 Share Posted April 13, 2008 So basically the only way voltage can drop is due to resistance isn't it. By [math]R=\frac{U}{I}[/math] we can see that resistance drops with the drop of voltage and with the increase of intensity. Link to comment Share on other sites More sharing options...
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