Timeline for Why do charges not lose potential as they travel through the circuit before reaching a resistor?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3, 2020 at 1:51 | vote | accept | Grant | ||
| Mar 3, 2020 at 1:14 | comment | added | Grant | Let me summarize what I am getting from u @BowlOfRed. so what you are saying is that there will be an accumulation of charges on the higher voltage side of resistor, which will essentially cancel out the potential of the battery? This charge will then start to flow down the resistor, and as charges leave and potential drops then more charge is added due to the E field of the battery? This means there will initially be a voltage drop across the wire but it will equalize fast? | |
| Mar 3, 2020 at 1:03 | comment | added | Grant | I added a picture above in the orignal post, it seems as though you are talking about when there is not a path for the charges to flow, but when there is a path there will be a field inside correct? that doesn't equal out until charge has moved from the cathode to anode (or anode to cathode) until the conductor is in static equilibrium. Kind of like the picture above. | |
| Mar 3, 2020 at 1:00 | comment | added | BowlOfRed | The net field inside is negligible because the charges in the wire react to the (external field) by moving in such a way that an opposing field is created. The battery's field is still contributing, but a cancelling field is also present. | |
| Mar 3, 2020 at 0:53 | comment | added | Grant | the potential at the top end of the battery due to the accumulation of charges. Also why do you say the field inside the wire is negligible? There is a field generated by the battery, and since the non induced E force is a conservative field there will be an electric field experienced inside the conductor regardless of direction. | |
| Mar 3, 2020 at 0:52 | comment | added | Grant | Okay let me elaborate, the problem I have there is that the electric field due to the anode of the battery is smaller at the point where the resistor is, so the accumulation of charges strictly to the combative electric field of the anode versus top end of resistor would lead to a smaller potential (smaller accumulation of charges) at the top end of the resistor. The charges moving with the drift velocity (kinetic energy) has momentum associated with it which push aganist the electric field of the accumulated charges at the resistor, therefore this allows more charges to accumulate, restoring | |
| Mar 3, 2020 at 0:50 | comment | added | BowlOfRed | Agree on the first 2 sentences. In the third, I'm not sure the KE "pushes" against anything. The KE of the charges is established in the circuit (and constant throughout), and the field inside the wire is negligible, so the charges move in the wire with no change in PE or KE. | |
| Mar 3, 2020 at 0:45 | comment | added | Grant | Okay, so essentially what I am seeing is that charges will lose potential as it is translated to kinetic energy, but when they reach the resistor they essentially hit a barrier or bump in which they are slowed down. Doing so we get an accumulation of charges. The kinetic energy pushes aganist the resistive force of the accumulated E field so that in general the potential at this point is the same as it was at the anode of the battery? | |
| Mar 3, 2020 at 0:34 | history | answered | BowlOfRed | CC BY-SA 4.0 |