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Let's imagine a hypothetical circuit where there are a large number of wires placed in parallel to each other, hooked up to a simple power source.

We know that voltage at each wire would be equal $V_{total}=V_1=V_2=...=V_n$ where $n$ approaches a large number; and that each wire is of some arbitrary constant length.

Next, assume that at the start of each wire there is a single charge of $+1C$, in each wire placed in parallel.
Since work done on a charge is $W=VQ$; where $W=$ work done, thus we apply the same voltage to each charge in each wire placed in parallel.

Since the voltage across each wire would be the same (say, $Resistance$ is ineligible, but $\neq0$) the work done would be same. Additionally, we know $W=\vec{F}.\vec{s}$; Since the charge is displaced to a significant length (i.e of the wire) Thus work is done even if we may not be able to easily quantify force.

My questions is this - if the number of parallely-placed wires increases, $W\uparrow$. Thus, we can gain infinite joules by placing more and more parallel wires violating the conservation of energy:

\begin{equation} \sum_{i=0}^{\infty}W_i = V_i \times1 \end{equation} by moving the $+1C$ charge in each parallel wire.

How is that possible?

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    $\begingroup$ "We know that voltage at each wire would be equal $V_{total}=V_1+V_2+...+V_n$". No. For a parallel circuit $V_{1}=V_{2}=V_{3}=...V_n$. $\endgroup$
    – Bob D
    Commented Nov 27, 2021 at 20:57
  • $\begingroup$ @BobD good catch, completely forgot about that. Thanks! :) $\endgroup$
    – neel g
    Commented Nov 27, 2021 at 21:03
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    $\begingroup$ "My questions is this - if the number of parallely-placed wires increases, $W\uparrow$. Thus, we can gain infinite joules by placing more and more parallel wires violating the conservation of energy:" I don't follow you. The energy is supplied by the voltage source. Why would there be a violation of conservation of energy as long as the voltage source could supply the energy? Perhaps I don't understand what you are getting at. $\endgroup$
    – Bob D
    Commented Nov 27, 2021 at 21:48
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    $\begingroup$ The premise is incorrect. Sure, you could transfer infinite energy with infinite branches, a perfect voltage source, and an ideal circuit, but none is possible in practice, so there’s no problem except a practical limitation to these idealizations. $\endgroup$
    – Chemomechanics
    Commented Nov 28, 2021 at 1:51
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    $\begingroup$ Work is performed by the source, but if that could violate conservation of energy then it would do so with any number of wires, even just one. The fact is that performing the work of moving charge requires the source to expend energy. There is no net gain of energy. This is why batteries are depleted by use, for example. $\endgroup$
    – John Bollinger
    Commented Nov 29, 2021 at 6:50

7 Answers 7

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For the same emf, as you increase the number of parallel circuit paths (i.e. increase $n$), you end up splitting the total charge $Q$ into smaller charges $q_i$ where the $q_i$s go through the parallel paths.

So, if you keep the same emf source and increase $n$, the charge $q_i$ that goes through a parallel path will be a fraction of the total charge $Q$ such that $\displaystyle \sum_i q_i = Q$ where $Q$ is the total charge passing through the emf.

To keep the charge that passes through each parallel path the same, you'd have to increase the emf as you increase $n$.

Either way, conservation of energy is not violated.

The reason is that for a parallel circuit, we have $I = I_1 + I_2 + I_3 +\dots + I_n$ and so $Q=q_1 + q_2 + q_3 + \dots + q_n$ which means that the work is $$\begin{align}W&=\sum_i^n W_i \\ &=\sum_i^n q_iV \quad \text{ and since }V\text{ is constant} \\ &=V\sum_i^n q_i\quad \text{where }Q=\sum_i^n q_i\\ \implies W&=VQ \end{align}$$

Suppose we represent the total charge $Q$ with these orange balls, notice how it is split into smaller charges $q_i$ which go through the parallel circuit such that the total work still is $VQ$.

enter image description here

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  • $\begingroup$ "For the same emf, as you increase the number of parallel circuit paths (i.e. increase $n$), you end up decreasing the charge that goes through each path". I don't understand what you mean. Are you saying the current decreases because of the internal resistance of the battery, which results in a drop in the terminal voltage? $\endgroup$
    – Bob D
    Commented Nov 27, 2021 at 21:58
  • $\begingroup$ @BobD I meant that the total charge charge that goes through the circuit is split. I should have used "split" and not decrease $\endgroup$
    – user256872
    Commented Nov 27, 2021 at 22:01
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The work done by the battery would have to be equal to (or greater than) your final formula.

The Work done by the battery is the E.M.F. of the battery multiplied by the charge passing through it, but many $1C$ charges would pass through the battery, $1C$ for each parallel wire.

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Since work done on a charge is $W=VQ$; where $W=$ work done, thus we apply the same voltage to each charge in each wire placed in parallel.

Be careful. The voltage $V$ or potential difference across each wire of the parallel circuit is the work required per unit charge to move the unit charge through the wire. But the total work is in moving the total charge through the wires. That will not be the same unless they all have the same resistance (which you did not specify) and therefore the same current (charge per second through each wire).

Since the voltage across each wire would be the same (say, $Resistance$ is ineligible, but $\neq0$) the work done would be same.

Only if each wire has the same resistance. (I assume you meant negligible and not ineligible).

Additionally, we know $W=\vec{F}.\vec{s}$; Since the charge is displaced to a significant length (i.e of the wire) Thus work is done even if we may not be able to easily quantify force.

But you can quantify the force. The force applied to each charge is

$$\vec F=Q\vec E$$

therefore

$$W=QEl$$

where $l$ is the length of the wire. If the wires have the same resistance, even if small, then $E=\frac{V}{l}$ is the same in each wire.

My questions is this - if the number of parallely-placed wires increases, $W\uparrow$. Thus, we can gain infinite joules by placing more and more parallel wires violating the conservation of energy:

I don't follow you. The energy is supplied by the voltage source. Why would there be a violation of conservation of energy as long as the voltage source could supply the energy? Perhaps I don't understand what you are getting at at this point.

Hope this helps.

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Yes, if you had a perfectly ideal voltage source capable of maintaining a constant non-zero voltage differential between its terminals, regardless of what's connected between them, then you could theoretically obtain an unlimited amount of power from it just by connecting more and more loads in parallel.

For example, let's say that your voltage source provides a constant voltage differential of 1 V, and you connect a 1 Ω resistor across it. Now, per Ohm's law, a current of 1 V / 1 Ω = 1 A will flow through the resistor, generating 1 V × 1 A = 1 W of heat in the process.

But if you now connect two 1 Ω resistors in parallel across the same ideal voltage source, then a 1 A current will now flow through each of them, generating 2 W of heat. And if you connected $n$ such resistors in parallel, then a 1 A current would flow through each of them, and you would get $n$ watts of heat output in total, for any number $n$.

Of course, in practice that won't happen, because there's no such thing as a perfectly ideal voltage source.

In fact, the thought experiment above shows exactly why such a perfect voltage source cannot exist: by connecting arbitrarily many parallel loads to such a source, we could obtain unlimited power from it.

What happens in reality, if you try to do that, is one of two things:

  • First, every real voltage source has (or behaves as if it has) some non-zero internal resistance, which causes its output voltage to drop as the current draw increases, limiting the maximum power the source can deliver to less than some finite number of watts. If you're lucky, the source can safely supply this much power, and nothing else happens.

  • If you're not so lucky, the source is not designed to be safely short-circuited like that, and will overheat or otherwise fail. If you're really unlucky, it might overheat so fast that it explodes or catches fire.

That second possibility, by the way, is why all the electrical wiring in your house goes through a fuse or a circuit breaker. The national power grid that your house wiring is connected to is just about the closest thing to an ideal (AC) voltage source you can find, and could, in theory, supply a huge amount of electrical power without any noticeable voltage drop — way more than the wiring inside your walls could handle without overheating and catching fire! Those safety devices are specifically designed to fail (reversibly, in the case of circuit breakers) and break the circuit if you try to draw too much power through them (e.g. due to an accidental short circuit), and to do so first before anything else overheats and fails in a more destructive manner.

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It sounds like you'd have a circuit like this:

 +----[ voltage source ]-----+
 |                           |
 +----[ resistor/wire ]------+
 |                           |
 +----[ resistor/wire ]------+
 |        ...                |

From the electrical engineering 101 standpoint, adding more wires just decreases the total resistance between the terminals of the voltage source. The power provided by the source is $ P = U I = U^2 / R $ and with an ideal voltage source, you could indeed get an arbitrarily large power output (and hence, energy), if you had an ideal voltage source.

But real voltage sources aren't ideal. Raising the load often leads to the voltage dropping, possibly the source overheating (due to internal losses) and eventually shutting down or blowing up. (Probably so that some internal part overheats and blows up, resulting in the whole thing shutting down.)

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each wire gets q/n charge (assuming each wire has equal resistance). Work done by battery is qV. Now since each parallel branch got q/n charge and there are n branches. The total work done by battery is n * [V * q/n] = qV. if u differentiate with respect to t on both side, you would get IV which is the power or energy consumed per unit time. Either you connect in parallel or the resultant resistance, the conservation law is not violated. The same potential across all branches does not alone decide the energy consumed by that branch, but also how many charges reached that branch.

Note: Voltage is across each wire in parallel is same, but the charges received in each branch changes as per the number of branches and resistance.

I hope this helps.

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It does not violate the conservation of energy rule. If you add more parallel wires instead of a single wire, the charge will get distributed between multiple wires but the total charge $Q$, which in second case will be the sum of all the smaller charges pushed through multiple wires, will stay the same. Attempting to drink a beverage through 10 straws at once, instead of 1 straw, won't make you empty the cup 10 times faster.

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