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I am trying to conceptualize the time which a battery will last in a given circuit. Below is an example from which I will explain the questions which I am struggling with : circuits

Lets begin by asking this question: In these circuits with identical batteries and bulbs, compare the time the power sources in these circuits will last . the answer is: both will last the same time. However, I solved this question by giving a value to current (i). In the circuit in the left, it is evident that if the voltage lost over the bulb is V, the current produced by the individual batteries will be (i/2) AND the time which the "power source" will last can be calculated by t=(E.E/P) . My question is regarding the second circuit. In naming the current, do we say that the individual batteries produce (i/2) current ? If the individual batteries produce (i/2) current, then shouldn't the time in which the "power source" finishes be different than the other circuit, since all batteries will finish at the same time.

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Batteries, ultimately, store electrical energy. They die when this electrical energy is depleted. As such, the relevant factor for battery lifetime is how quickly this electrical energy is depleted; in other words, the power $P=VI$ produced by the battery, for potential difference $V$ across the battery and current $I$ through the battery.

(Ideal) batteries produce power by creating a fixed potential difference; in other words, they are voltage sources. The current in the circuit depends on the resistance of other components. Suppose that each bulb has a resistance $R$. In the first circuit, each battery produces a voltage $V$ across its individual branch, meaning that the bulb experiences a potential difference $V$. Therefore, the current flowing through the bulb is $I=\frac{V}{R}$. This current splits equally between the two branches, each containing one battery, so each battery has a current $\frac{I}{2}$ passing through it. Therefore, the power output of each battery in the first circuit is $P=V\frac{I}{2}=\frac{1}{2}\frac{V^2}{R}$.

In the second circuit, we have two batteries in each branch, meaning that each branch produces a potential difference of $2\cdot V$. Each bulb has a resistance of $R$, so the two bulbs in series have a total resistance of $2R$. Therefore, the total current through the bulbs is $I=\frac{2\cdot V}{2R}=\frac{V}{R}$. Once again, each branch receives half of the current flowing through the bulbs, and once again, the potential difference across each battery is $V$, so the power output of each battery in the second cricuit is $P=V\frac{I}{2}=\frac{1}{2}\frac{V^2}{R}$. Since the power consumed by each battery in the two circuits is the same, their lifetime should also be the same.

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  • $\begingroup$ Just FYI, on the first read, I thought the first sentence of the last paragraph stated that the voltage across each branch was 2 volts rather than 2 times the voltage $V$ so I've edited that for clarity. $\endgroup$ – Alfred Centauri Oct 29 '18 at 18:10
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the answer is: both will last the same time

Here's an (I think) interesting alternative path to the answer.

Due to the symmetry of the right-most circuit (assuming identical cells and bulbs all around), the junction of the bulbs and the junctions of the cells have identical potential and so, one can connect these junctions together with a wire.

Then it's clear that the right-most circuit breaks down into two copies of the left-most circuit that are connected together at a single node which is to say that we could separate the two copies without changing the voltage across or current through any circuit element.

From this, conclude that the right-most circuit 'runs down' in the same amount of time as the left-most circuit.

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