When two resistors with equal resistances are wired in series the voltage across each resistor is equal to half the voltage of the power supply. Is this also the case the instant the power supply is turned on? At the instaneous moment when the electric current has just reached the first resistor in series is the voltage across that resistor equal to half the voltage of the battery or all the voltage of the battery? In the case of the former being true, how is the current "aware" of the second resistor in the circuit before encountering it? In the case of the later, how does the circuit eventually reach equilibrium?
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1\$\begingroup\$ Effectively, yes. If a charge moves in at one end then one is pushed out at the other end. The speed of propagation of the EM (electromagnetic) wave is close to the speed of light. For most low frequency applications we just consider that the current is instantaneously the same through all points on that branch of the circuit. \$\endgroup\$– TransistorSep 17, 2017 at 17:35
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\$\begingroup\$ It sounds as though you want to know how it is that charges are aware of "how to move" in a circuit. So, for example, how is it that a current of charges "knows" to split in a certain proportion when they reach a Y in a wire, dividing up so that some portion moves down one path and a different portion moves down another path, as it is impossible for them to "know" what's ahead. Is that it? \$\endgroup\$– jonkSep 17, 2017 at 18:00
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\$\begingroup\$ @Transistor unless of course the wires between the resistors are very very very long. Then you may get into considerable measurable transmission line effects.. \$\endgroup\$– Trevor_GSep 18, 2017 at 18:32
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\$\begingroup\$ @jonk Yes, that is essentially my question. It seems as though the current is able to foresee the path it is going to take. \$\endgroup\$– JgbcodeSep 18, 2017 at 18:42
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\$\begingroup\$ @Jgbcode I think the best way to figure this out would be to get and read Chabay & Sherwood's "Matter & Interactions", 3rd edition (or later?), and pay a lot of attention to Chapter 19 there. Also, you will learn from it the difference in meaning between "equilibrium" and "steady" states, as they use these terms advisedly there. Are you willing to accept a simplified answer? (The book is better than I would be, so if you'd prefer to get the book that is fine with me.) +1 as now this is a good question, I think. \$\endgroup\$– jonkSep 18, 2017 at 21:10
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Is this also the case the instant the power supply is turned on?
For ideal resistors in circuit theory, it's true as soon as the power supply is turned on.
In the real world, no power supply actually turns on instantly. And each resistor has a small parasitic capacitance. And the wires connecting the power supply to the resistors have parasitic indcutance. And each wire has a parasitic capacitance to each other wire in the circuit. If you introduce the effects of these parasitics into your ideal circuit model, you will find that the current does not equalize perfectly through the main wires because some goes through the parasitic capacitors. But this effect is miniscule, due to the non-zero turn-on time of the supply and the wire inductances slow the current turn-on and allow it to stay pretty darn near equal around the loop as it turns on.
answered Sep 17, 2017 at 17:38