# Example: Batteries in series and parallel

## Understanding the situation

In our previous two examples of how to use Kirchhoff's principles to analyze electrical networks, resistors in series and resistors in parallel, the analysis was fairly straightforward.

In the series case, the two resistors acted as a single effective resistor with a resistance equal to the sum of the resistances of the individual resistors. In the parallel case, the two resistors acted as a single effective resistor equal where the sum of the reciprocal resistances yielded an effective reciprocal resistance.

That's because resistors in series are more difficult to push a current through — the current has to go through both resistors — and resistors in parallel are easier to push a current through — the current can split between the two resistors. So both series and parallel resistors are in some sense equivalent to the simplest case of one resistor connected to one battery.

But lots of important electrical situations in biology can't be modeled by such simple systems. For example, what happens in a cell membrane (especially a nerve cell) is a highly electric phenomenon, depending on charges flowing through the membrane. Ion pumps act as batteries to create potential differences, and since there are multiple ions pumps that can pump ions either into or out of cells, a one-battery model doesn't suffice. Also, looking at models with more than one battery gives us a better insight into how Kirchhoff's principles actually function in more realistic circuits and shows their value in interpreting more complex situations.

To see how this works, lets solve a toy model problem with two ideal batteries in series and two ideal batteries in parallel.

## Presenting a sample problem

Consider the three electrical networks shown at the right. Identical batteries are connected in different arrangements to the same light bulb. Assume the batteries have negligible internal resistances. The positive terminal of each battery is marked with a plus. Rank these arrangements on the basis of bulb brightness from the highest to the lowest.

## Solving this problem

The brightness of a bulb is proportional to the power it dissipates (the Wattage). If we move a small bit of charge $dq$ through a resistor so that it experiences a voltage difference $ΔV$, we do work $dq ΔV$. If we do this in a time $dt$, the rate at which energy is used is $dq/dt ΔV$ or

$$P = I ΔV$$

the product of the current through the resistor times the voltage drop across it. If the resistor is Ohmic, we can write $ΔV = IR$ so we could also write the power as $P = I^2R$.

This tells us that if the bulb converts all the electrical energy it dissipates in its resistance as light (not really correct for an incandescent bulb that gets hot, but better for an LED) the brightness of the bulb will go as the square of the current through it. So let's find the current in each of the bulbs for the three cases.

Let's go through the cases one at a time.

#### Case A: One battery

Case A is simple: just a single battery connected to a single resistor. Kirchhoff's current rule tells us that the same current has to go through everything. Let's call it $I$.  Choose the low end of the battery as our 0. Then the high end of the battery is at $V_0$ and our resistanceless conductor heuristic let's us map the potential everywhere as shown. We see that the voltage drop across the resistor is $V_0$ and the current through it is $I$. Ohm's law relates the voltage drop across the resistor to the current through it. In this case we can assume we are given the voltage rise for the battery, $V_0$, and the resistance of the bulb, $R$. We want to find the current through the bulb. Expressing our current in terms of our given parameters, we just get

$$I = \Delta V/R = V_0/R$$

$$P = \frac{V_0^2}{R}$$

#### Case B: Batteries in series

Now let's consider case B.  We still only have one loop so whatever goes through one element must go through them all by Kirchhoff's loop rule.

Now let's map the potential. We can choose one 0 for the potential so let's choose it as the low end of the right battery. We can continue those 0's down the right side as we did before, but as we go to the left something different happens. The high side of the left battery is connected to the low side of the right battery! So we can't take the low side of the left battery as 0: it has to be at a potential V0. But we know a battery rises the potential from its low terminal to its high by a fixed amount, whatever the low terminal's value is. So if we rise by V0 going across the left battery, we must wind up at a potential of 2V0! By our resistanceless conductor heuristic we have to have this same value all along the wire until we get to the resistor.

So the drop across the resistor in this case is $ΔV = 2V_0$. Ohm's law gives us the current in the resistor and as a result, the power:

$$I= \frac{ΔV}{R} = \frac{2V_0}{R}$$

$$P = I^2R = \frac{4V_0^2}{R}$$

Two batteries in series makes the bulb use four times as much power as one! Doubling the voltage in this arrangement both doubles the voltage drop across the resistor and the current through it. The bulb will be much brighter.

#### Case C: Batteries in parallel

Now let's consider case C. This is a bit trickier since there are two loops. We only care about the current through the bulb (the resistor) so let's call the current through that loop $I$. After passing through the resistor, when it reaches the first split, the sum of the current out of since the two possible paths have to add to I by Kirchhoff's current rule. Since the two paths it can take are identical, we assume it splits in half, as shown. (Note: the labels on the current on the upper loop mean "current divided by 2" not "one-half".)

Now let's do the map of the potential. We can choose one point in the network as the 0 of our potential. Let's take it as the low end of the lower battery. By the resistanceless conductor principle, any part of a wire connected to that point with no resistance between them must also be at 0. This means the low end of the second battery and the right side of the resistor. (Follow the wires from the low end of the battery and see all the places we have marked as 0.)

The principle that a battery creates a rise of potential of a fixed value across its terminals means that the high side of both batteries must be $V_0$. Using the resistanceless conductor principle, we can find the values of the potential everywhere else in the circuit, including at the other end of the resistor. This tells us that the potential drop across the resistor is $V_0$. By Ohm's law (Kirchhoff's resistance principle) we can relate the current to the drop across the resistor:

$$I= \frac{ΔV}{R} = \frac{V_0}{R}.$$

This is the same result as we had in case A, so our brightness ordering for the three cases is

B > A = C.

## Interpreting the results.

The two batteries in series provide twice as much "push" (electrical tension) as one, so we get more current and more power dissipated in the bulb. The two batteries in parallel provide the same voltage drop, so our bulb won't be brighter, but since the current through each battery is halved, the batteries will last longer.

While this solves our problem, it is worth looking a bit at case C a bit more. We assumed the batteries are identical, but suppose they weren't? We would have gotten into trouble! We'd have run our 0 values of $V$ up the right side and then jumped across each battery with their appropriate rises and gotten different values for what the potential should be down the left side of the network. A contradiction! The reason for this is we made the assumption at the beginning that our batteries were ideal: that they always maintained a fixed voltage difference across their terminals and had no internal resistance. We also assumed that our wires were ideal: had 0 resistance.

These assumptions are fine as long as there are always resistors in the circuits that are significantly bigger than the (usually small) resistances of the batteries and wires. But in the upper loop of case C we have connected two batteries together in a loop without any resistors. This is like connecting both terminals of the battery to each other with just a wire — a short circuit. Since  $ΔV$ is fixed and $R ~ 0$, we get $I = ΔV/R$ is very large. The large current means a lot of power dissipated in the very small resistance of the wire. It gets very hot and drains the energy of the battery very fast. Don't try this at home!  (Or in your lab!) You may not start a fire, but you will almost certainly drain your batteries.

Joe Redish 3/15/16

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