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This diagram problem has got me thinking about what determines the voltage in a battery. At school, we were taught this method of giving a value of voltage to points in a circuit. However, I do not understand how we add up the voltages and I also don't understand why the voltage between the two batteries is V and not 2V .It seems to me that the overall voltage across the circuit should be 2V. Is my idea of viewing Voltage of batteries as fields wrong? How is the voltage of batteries determined? What properties of the battery allow the voltages to be "added up" as in the diagram?

The reason why I believe that voltage is a field is because the potential distance is established in a circuit in almost the speed of light. Whereas, particles do not move in such a speed in the circuit. This is why the addition done here bothers me. here is the circuit

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  • $\begingroup$ In the water analogy, voltage corresponds to the height of the water. You can therefore view a voltage source as a pump that lifts up the water by some certain amount. Having two voltage sources first 'lifts' the water up to one level (1 V), and then to the second level (2 V). The water can then flow to ground level through the resistor, which 'drops 2 V' over it. $\endgroup$ – ahemmetter Oct 31 '18 at 8:16
  • $\begingroup$ The voltage of an individual, ideal battery in a circuit diagram is whatever we declare it to be. The voltage of an individual, practical battery in an actual electronic device is determined by chemistry, and if you want to know more, chemistry.stackexchange.com probably is the place to ask. $\endgroup$ – Solomon Slow Oct 31 '18 at 13:42
  • $\begingroup$ "Whereas, particles do not move in such a speed in the circuit." - this is true but why does that matter? The conductors connecting the circuit element are 'full' of mobile electrons. You're not thinking that the conductors must slowly 'fill-up' with electrons from the cell(s) in order to establish a steady current are you? $\endgroup$ – Alfred Centauri Oct 31 '18 at 14:41
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Different points in the gravitational field have different potentials. If we place an object in a point with high potential, it has more stored energy than if it's placed in a point with low potential. In the zero point, the potential i zero. This way of thinking is also applicable on an electric field, where there in a similar way are diffrent points with different potentials. Kirchhoffs second law states that the sum of the potential differences (voltage) around a closed circuit is zero. This is no other way than saying that if you go from your home (zero point) to the grocery store and back home (zero point), the sum of the potential differences will always be zero no matter what way you decide to take.

To get to the grocery store you first have to climb a mountain with the height h. Once you have reached that height, your realize that there's another mountain that is h high on top of the mountain you have just climbed. After you have climbed a total of 2h, you reach the grocery store where you decide to buy a parachute. You then use your potential energy and fly down to the bottom and walk home again on flat ground.

mgh (first mountain) + mgh (second mountain) - mg2h (parachuting) = 0.

This adventure is analogous to your circuit. Your home is at "0" and the grocery store to the left of both batteries - your walk follows the direction of the current.

v (first battery) + v (second battery) - 2v (resistor) = 0.

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  • $\begingroup$ Welcome New contributor Oscar! Would you explain why the voltage across the resistor is $3V$ when each cell has (I believe) $V$ volts across? $\endgroup$ – Alfred Centauri Oct 31 '18 at 14:20
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    $\begingroup$ Thanks @AlfredCentauri. I misunderstood the drawing, you are correct. The drop is 2V. $\endgroup$ – Oscar Oct 31 '18 at 14:42
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It's my understanding that it's fixed by the various potentials & stuff of the chemical species whose chemical reactions generate the EMF.

Whilst on this subject, I'd like to mention how in electronics teaching a capacitor is often compared to a battery. I would say it's very misleading to do so; and the comparison really ought not to be broached at anyone who is learning electronics in earnest. In a capacitor the voltage is proportional to the charge on it, and the constant of proportionality is the reciprocal of the capacitance. A battery (or cell, rather) is a sort of engine in series with an internal resistance: the effect of the running-down of the cell is not to reduce the force delivered by this engine, but rather to reduce the quantity of stuff it can shift, and this is manifested externally as an increase in the internal resistance. For an elementary theoretically perfect cell, if it is literally capable of supplying only a single electron more, that electron will be supplied at the nominal voltage.

By elementary, I mean something like a zinc rod & a copper rod dipped in some dilute hydrochloric acid. A single molecule of HCl would undergo the reaction with the same potentials figuring as would for any one of many molecules undergoing it simultaneously ... provided the solution is dilute. The voltage is determined by the difference in electronegativity beween copper & zinc. Real commercial cells do show a considerable drop in open circuit voltage as they run down, though, as there is much more to them than just two different metals dipped in acid, and the reactants are highly concentrated; but it's just a bit of sagging towards the low-charge end of the curve - nothing remotely like the direct proportionality of the capacitor.

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At school, we were taught this method of giving a value of voltage to points in a circuit.

I believe there's a misconception here that may be the root of the remainder of the questions (of which there are too many for one post).

I assume that, by "voltage to points in a circuit", you're referring to the concept of a node voltage but, conceptually, a node voltage is not a 'voltage at a point in a circuit'.

Rather, a node voltage is the potential difference between that node and the 'ground' (reference, datum) node.

The node voltage of the reference node is zero since the potential difference between a node and itself is zero.

For a physical picture, imagine placing the black lead of your voltmeter on the reference node and then placing the red lead there too. Your voltmeter should read zero volts.

Now, place the red lead at a different circuit node. The measured voltage is the node voltage for that node. Again, this is not a voltage at a point in a circuit but the voltage between that node and the reference node.

However, I do not understand how we add up the voltages and I also don't understand why the voltage between the two batteries is V and not 2V

So, for example, if I place the black lead of my voltmeter at the reference node ( marked with $0$ in your diagram), and I place the red lead of my voltmeter at the node between the two cells, I should of course measure the voltage $V$ across the battery. That's why that node has voltage $V$ assigned; it is the voltage between that node and the reference node.

Now, assuming identical cells, the voltage across the left-most cell is $V$ and that means that the $+$ terminal is more positive by $V$ than the $-$ terminal. But we already know the $-$ terminal is connected to the node with node voltage $V$ thus, the node voltage at the $+$ terminal is $V + V = 2\cdot V$.

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