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The way I understand voltage is the following: Its a measure of energy given to charge carriers in a circuit. Its a measure of potential difference, and it is the work in joules required to move 1 coulomb of charge from one point to another. A batteries voltage effectively pushes electrons around and gives them energy to do so. But I can't seem to understand why the total voltage drop in a circuit is equal to the voltage of the battery? I the battery gives electrons this energy, where does it go, otherwise wouldn't electrons continuously gain energy?

Further If we had a circuit with just a battery and wires connecting the negative and positive terminal, would there be a voltage drop? What would happen and why?

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But I can't seem to understand why the total voltage drop in a circuit is equal to the voltage of the battery?

Voltage represents a potential difference, just like the height of the hill represents a potential difference. If you hike up a hill you gain potential energy that's ONLY a function of the height regardless which way you take.

I the battery gives electrons this energy, where does it go, otherwise wouldn't electrons continuously gain energy?

The electrons do indeed gain energy. They get accelerated and bump into the crystals of the conductive material which generates heat. The amount of heat is directly proportional of the voltage difference and the number of electrons per second.

Further If we had a circuit with just a battery and wires connecting the negative and positive terminal, would there be a voltage drop? What would happen and why?

DO NOT EVER DO THAT. That's how fires get started and people get hurt or killed.

Yes, there will be a voltage drop but since the wire has very low resistance the amount of electrons is only limited by the internal workings of the battery. The battery will release it's entire internal energy in a very short time and depending on the type and size of battery, this can result in serious damage. The wire may melt, the battery may cook and/or explode, nearby objects can catch on fire, etc.

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  • $\begingroup$ Voltage in a circuit is analogous to height in a gravitational field. If you hoist a mass up to some height and drop it back to the ground, it will fall as far as you hoisted it. $\endgroup$
    – John Doty
    Commented Jul 9, 2022 at 11:56
  • $\begingroup$ @John Doty If I had a circuit with just a lightbulb and cell, then I measure the voltage between the bulb and cell it gives me 0. But when I measure the voltage between both ends of the bulb it gives me the battery voltage. If the voltage was 0 for the first one, then how do electrons even move if there is no potential difference? $\endgroup$ Commented Jul 9, 2022 at 20:03
  • $\begingroup$ Shouldn't the voltage be the same across any two points in a circuit. Why is some measured voltage 0? $\endgroup$ Commented Jul 9, 2022 at 20:04
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The E field is path independant.

If the voltage across the plates is 5v, then any path with the same start and end points as the plates, will also give 5v

This is because $\nabla × \vec{E} = 0$ for typical circuits.

In wires there's resistance, the electrons keep at a constant velocity, this other energy is transferee to the lattice as heat.

There is a PD drop in a circuit containing a battery and a wire, otherwise there wouldn't be an E field

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But I can't seem to understand why the total voltage drop in a circuit is equal to the voltage of the battery?

This is based on Kirchhoff's voltage law that the algebraic sum of the voltages around a closed loop is zero. It is a requirement for conservation of energy. See below.

If the battery gives electrons this energy, where does it go, otherwise wouldn't electrons continuously gain energy?

The electrical potential energy given the charge by the battery is either dissipated as heat in resistance (due to collisions between the charge and the particles of the resistance), or stored as energy in the electric and magnetic fields of any capacitance or inductance in the circuit.

Further If we had a circuit with just a battery and wires connecting the negative and positive terminal, would there be a voltage drop? What would happen and why?

With the exception of superconductors, all conductors (including wires) have some resistance. The voltage drop $V$ across the wires connecting the positive and negative terminals would simply be, from Ohm's law, $V=IR$, where $R$ is the resistance of the wires.

It should be noted that the voltage drop $V$ across the wires is not the voltage between the battery terminals before the wires are connected, but after. That's because all real batteries have internal resistance which effectively limits the maximum amount of current a battery can deliver. So the current delivered by the battery is

$$I=\frac{emf}{r_{b}+R}$$

Where $r_b$ is the internal battery resistance and $emf$ is the battery voltage with nothing connected to the terminals.

Hope this helps.

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