Why is current constant through a resistor?

[BuggyWungos]

Homework Statement

"Consider the circuit shown. When hooked up to a certain battery, there will be a current, I, moving to the right in the top wire (above resistor A). How would the current through resistor A compare to the current through the bottom section of wire between the points marked n and m?" (see image below)

Relevant Equations

$$\delta V = IR$$
$$I_a=I_b+I_c$$

I understand that the current in the top wire above A will be the same as the wire just below A, as it will be for the wire region marked by n to m, due to the fact that all current in equals all current out.

My confusion lies in the first equation I supplied, the change in voltage through the resistor A should be zero, and if you rework the equation a little bit, the current should drop as it goes through resistor A, but that doesn't seem to be the case, and it also wouldn't explain why the current goes back to it's original value after leaving A.

My understanding is as goes:
1) Voltage drop ($\delta V$) through A will be zero, no change in voltage, meaning current should be 0?
2) If there was a non-zero voltage drop through A, then current through A will be less than current above A?

 

[Orodruin]

Your assumption 1 is incorrect. So is your assumption 2.

If the current through A was any less than through the wires connecting to A, then there would be charge accumulation in A.

 

[kuruman]

and if you rework the equation a little bit, the current should drop as it goes through resistor A

Why should it? When you write $~\delta V = IR~$, $\delta V$ is the potential difference across the entire resistor, $I$ is the current through the resistor and $R$ is the constant resistance. The current or charge through the resistor per unit time, is the same as long as the other two quantities do not change. What goes in must come out. The only quantity that drops across the resistor is the voltage and that is why the product $IR$ is often called "the voltage drop".

 

[BuggyWungos]

Why should it? When you write $~\delta V = IR~$, $\delta V$ is the potential difference across the entire resistor, $I$ is the current through the resistor and $R$ is the constant resistance. The current or charge through the resistor per unit time, is the same as long as the other two quantities do not change. What goes in must come out. The only quantity that drops across the resistor is the voltage and that is why the product $IR$ is often called "the voltage drop".

Say we doubled the resistance in the resistor A, wouldn't the formula spit out current to be half that of what it originallly was due to $\delta V$ remaining constant? $$R_2 = 2R_1$$ $$\dfrac{\delta V}{2R_1} = \dfrac{1}{2}I$$

 

[kuruman]

Say we doubled the resistance in the resistor A, wouldn't the formula spit out current to be half that of what it originallly was due to $\delta V$ remaining constant? $$R_2 = 2R_1$$ $$\dfrac{\delta V}{2R_1} = \dfrac{1}{2}I$$

Yes, that is Ohm's law. What does this have to do this question? The fact remains that whatever current goes through resistor A at the top has to exit through wire section nm at the bottom. If you doubled resistor A, the current through it would be less than before but still that same current has to exit through wire section nm at the bottom.

 

[Orodruin]

Say we doubled the resistance in the resistor A, wouldn't the formula spit out current to be half that of what it originallly was due to $\delta V$ remaining constant? $$R_2 = 2R_1$$ $$\dfrac{\delta V}{2R_1} = \dfrac{1}{2}I$$

If you double the resistance of A in the circuit you will change the potential across A as it is in series with the BC parallel coupling. The current will therefore be something larger than half the current, but less than it was originally. This doesn’t change the fact that the current is the same as that going in and out of the circuit - that current also changes if you change A.

 

[BuggyWungos]

Yes, that is Ohm's law. What does this have to do this question? The fact remains that whatever current goes through resistor A at the top has to exit through wire section nm at the bottom. If you doubled resistor A, the current through it would be less than before but still that same current has to exit through wire section nm at the bottom.

Well, if that's true, why doesn't current get smaller as it goes through Resistor A compared to the wire above? If you applied Ohm's law to the wire above, you'd get a much larger current than applying ohms law on the resistor, in fact it would be infinite current... I know something isn't quite right here, but I'm not sure what it is.

 

[Orodruin]

Well, if that's true, why doesn't current get smaller as it goes through Resistor A compared to the wire above? If you applied Ohm's law to the wire above, you'd get a much larger current than applying ohms law on the resistor, in fact it would be infinite current... I know something isn't quite right here, but I'm not sure what it is.

The voltage drop in the (idealised) wire is zero so you cannot apply that reasoning. Even if the wire is not ideal the voltage drop will be much much smaller.

You try to apply Ohm’s law where you should be using Kirchhoff’s current law.

 

[jbriggs444]

Well, if that's true, why doesn't current get smaller as it goes through Resistor A compared to the wire above?

This sounds like a "leaky hose" model of resistance. Current does not leak out the sides of the resistor.

That is not how resistance works.

Try using a "partially clogged" model instead. What goes in must come out.

 

[BuggyWungos]

This sounds like a "leaky hose" model of resistance. Current does not leak out the sides of the resistor.

That is not how resistance works.

Try using a "partially clogged" model instead. What goes in must come out.

I'm going to simplify my model a little bit as I feel like I'm going in circles trying to understand what exactly is going on.

1) For an example like the one below, if I had thrown on a resistor with resistance any real number, the current through the resistor should be the same current throughout the wires as well?

2) If I take replace the resistor with another resistor that has infinite resistance, the current through the resistor will be 0, as is the current through the wires?

3) If I remove the resistor, the total circuit current will rise up, and I'll call this current $I_o$. If I then add in a resistor like in 1), the total current throughout the circuit will decrease to $\dfrac{\delta V}{R}$? In other words, no matter the situation (in this particular diagram), the current through the resistor will be the current through the entire circuit?

 

[Orodruin]

1 Yes

2 Yes

3 If you remove the resistance from the circuit, that’s a short circuit and the current is infinite. (In the ideal circuit. In a real circuit, the wires have some low but finite resistance)

 

[BuggyWungos]

1 Yes

2 Yes

3 If you remove the resistance from the circuit, that’s a short circuit and the current is infinite. (In the ideal circuit. In a real circuit, the wires have some low but finite resistance)

Right, cause $R=0$ and $\dfrac{\delta V}{R}$ would just be infinity :D

To finish it off, in my original example, $\dfrac{\delta V}{R_A} = I_A$ and $I_A$ should be equal to the current running through the wires above and at n-m?

If that's true, could I also say that the circuits current is $\dfrac{\delta V}{R_{eq}} = I_A$?