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Say you have a 9 volt battery, and you connect the two ends of the battery with a wire to form a circuit with no resistor. If you measure the potential difference between the two ends of the battery, the difference will obviously be 9 volts. This suggests that as you move along the wire, the potential slowly decreases incrementally. Now say you have the same battery and circuit, but you add a resistor. I've been told that the voltage drop across this resistor will be 9 volts. But this implies that now, as you move along the wire, there is no incremental change in voltage; voltage is constant until you get to the resistor, and then it drops 9 volts, and then it remains constant until you get to the other end of the circuit. This seems wrong to me, because if there is no voltage change in the wire, there must be no electric field, so what moves the electrons along? On the other hand, the drop in voltage across the resistor along with the resistance should determine the current in the circuit, so if the voltage drop across the resistor is not directly linked to the voltage drop in the battery, there would be no way to find the current, which also seems wrong. Unless, of course, we just approximate that most of the voltage drop occurs across the resistor. But in that case, are there any cases where that approximation does not work/ why is most of the voltage drop across the resistor?

In summary, I think that conceptually, I'm being tripped up as to why the simple act of adding a resistor to a circuit would supposedly cause the voltage change to occur at an isolated location (the resistor) and not over the whole wire. I'm just starting to learn about circuits so any insight would be helpful, and apologies if this is a trivial matter.

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If you measure the potential difference between the two ends of the battery, the difference will obviously be 9 volts.

Re-think your assumptions. A 9 V battery will not output anywhere near 9 V if you connect its terminals with a wire thicker than a strand of hair.

I've been told that the voltage drop across this resistor will be 9 volts. But this implies that now, as you move along the wire, there is no incremental change in voltage; voltage is constant until you get to the resistor, and then it drops 9 volts, and then it remains constant until you get to the other end of the circuit.

When you add the resistor, the current in the circuit will drop dramatically. So you should expect the voltage dropped by the wire to drop dramatically as well.

Generally it's a good approximation that the voltage dropped by the wire is negligible compared to the voltage dropped by the wire. In reality, there will be some voltage drop along the wire.

If you know what the wire is made of, you can look up that material's resistivity. Then if you know the wire's diameter and length, you can calculate its resistance. Then you can use the voltage divider rule to work out how much of the 9 V source voltage is dropped by the wire and how much by the resistor.

It wouldn't be uncommon for the wire resistance to be on the order of 1 - 100 milliohms. If you're comparing the voltage drop across this wire with the voltage drop across a 1 kilohm resistor, will it change your analysis much to simply ignore the effect of the wire?

But in that case, are there any cases where that approximation does not work/ why is most of the voltage drop across the resistor?

Yes there are cases where the approximation doesn't work. If we have a high-current load and we want to power it through the lowest cost (thus thinnest) wire, we need to determine how much voltage drop is acceptable across the wire, and choose the wire appropriately with this in mind.

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  • $\begingroup$ I understand most of this, and it mostly makes sense. But why is it not true that the potential difference between the ends of the battery is just 9 volts when a wire connects the two ends? I thought that by definition, a 9 volt battery has a potential difference of 9 volts between the two ends? $\endgroup$ – Marcel Mazur Oct 17 at 16:36
  • $\begingroup$ @Marcel, See What makes the internal resistance of a battery at microscopic level? $\endgroup$ – The Photon Oct 17 at 16:43
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    $\begingroup$ An ideal voltage source has a fixed voltage across its terminals by definition. A battery is a physical device that produces a voltage due to a chemical reaction, but the rate of that reaction is limited so the battery can't produce infinite current. As the current out of the battery increases, the voltage across its terminals decreases. And this effect is particularly strong in 9 V batteries, because they're made with 6 tiny 1.5 V cells connected in series. $\endgroup$ – The Photon Oct 17 at 16:45
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The voltage does indeed reduce gradually over the entire circuit, including the wires to and from the resistor.

In a simple series circuit the voltage drop over each of the elements (ie wires and resistors) in the circuit is proportional to their individual resistance. In power circuits the supply wires are designed to have a low resistance compared with the rest of the circuit they are supplying.

As a result, the voltage drop across the wires supplying power to a circuit is often neglected in simple calculations.

The approximation breaks down whenever the resistance of the supply leads is no longer negligible compared with the resistance of the rest of the circuit.

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