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Name: Andrea
Status: Educator
Grade: 9-12
Location: Outside U.S.
Country: United Kingdom
Date: Summer 2009

Why, in a DC circuit that has two batteries, does removing one of the batteries not half the current?

First let us talk about some background about the relationship between resistance, voltage, and current, and battery connections, then we will do detective work to determine why the current did not change.

Current in a DC circuit follows Ohm's law, V = I*R which says that the voltage "V" applied across a load equals the current "I" in amps multiplied by the resistance "R" of the load in Ohms.

Batteries can be connected in series, where the voltages add and the capacity is the same as that of a single battery, or in parallel, where the capacities add but the voltage stays the same. In series connection, the negative of one battery is connected to the positive of another. In a parallel connection, the positive connects to the positive and the negative to the negative. You can Google diagrams for series and parallel connections...I think the pictures make it a lot more obvious what the differences are between them.

If you removed a battery and the current did not change, then they must have been connected in parallel. If the resistance of the load did not change, then the only way the current would change is if the voltage from the battery changed. Since it did not change, then you must have had the batteries wired in parallel. If they had been in series, removing one would reduce the voltage (and, by Ohm's law, the current) by half.

David Brandt

I am assuming that the question is about an electric circuit that gets its power from two batteries placed in series. You are asking: if powered with only one battery (half the voltage), why is the current not one-half also?

For some simple circuits, the current will be exactly one-half. Specifically, if the circuit is a simple linear resistive circuit. In this case, doubling the voltage would double the current and so on in a "linear" fashion. You will learn about this in circuit class.

But not many practical circuits are simple linear resistive, especially if they contain transistors, capacitors, or inductors. These components are usually necessary to make a useful circuit. If the circuit makes power for a motor or lamp then the voltage/current relationship can be non-linear also.

Circuits can be designed with all sorts of relationships between voltage and current. For example, some standard circuits (which you can make with a handful of parts) draw a constant current regardless of the driving voltage. Or other circuits may double the current with half the voltage - this is useful for "constant power".

But most practical circuits are designed to do something useful, and are made up of many kinds of parts, and the voltage/current ratio is usually not "linear".

Robert A. Erck

My first guess is that the batteries are in parallel. In this case the voltage applied would not change. Therefore - assuming the resistance has not changed - the current would stay the same. This is assuming that the current is within the capacity of the remaining battery.

If the batteries were in series, removing one from the circuit would diminish the applied voltage and therefore diminish the current if the resistance remained unchanged.

Batteries in series provide greater applied voltage. Batteries in parallel will increase the possible current and increase battery life.

Larry Krengel

Hi Andrea,

If your DC circuit is just a simple resistance, then removing one of the two batteries will indeed halve the current. Ohm's Law states that I = V/R, that is, current through a circuit is equal to the voltage "V" that is driving the current, divided by the specific resistance "R" of the circuit. So if you had a circuit with a constant resistance, and you removed one of the two batteries, the voltage will be halved, and current flow would drop to half of its previous "2- battery" value.

In real life, things are slightly more complex. Batteries also have an internal resistance, which must be added in with the resistance of your circuit. So removing one battery, also removes that battery's internal resistance, and hence the total circuit resistance also changes. Normally, the internal resistance of a battery is very small compared to the resistance of the external "load", so internal battery resistance would likely not affect your results enough to worry about.

The main reason for the current through your circuit not dropping by exactly half when you eliminate one battery, is almost certainly caused by the load resistance in your circuit changing. A good example of such a resistance is a light bulb whose resistance changes dramatically with increasing brightness. Increasing the voltage causes the filament to get hotter (and burn more brightly), and this increased filament temperature causes its resistance to rise dramatically. So, doubling the voltage to a light bulb causes its resistance to rise, and since the bulb's resistance has increased, the current through it will not double (even though you have doubled the voltage that you drive the light with).

Electronic circuits (and even simple LEDs) are another example of devices whose resistance changes with driving voltage. However, if you are using an actual resistor as a load, then its resistance is constant no matter what driving voltage is used, and removing one of your two batteries will (almost!) halve the current.... the "almost" is the error caused by eliminating the internal resistance of one battery when you remove that battery.


Bob Wilson


You can connect batteries in two ways, in serial (one after the other) or in parallel (all of their positive terminals connected to each other and all of their negative terminals connected to each other.

For this problem, we will only consider fresh batteries that are at their full charge. That is, no old batteries that have lost half or three-quarters of their charge.

Please see the following URL for drawings of serial and parallel connected batteries:

The relevant equation is Volts (V) = Current (I) x Resistance (R).

In the top drawing the voltages of the batteries in series ADD to 8 volts. If we take a two ohm resistor and put the wires at the end of the string of batteries across it, we will get (8 Volts /2 Ohms) = 4 amperes of current. Take one of the batteries out of the string and re-connect the wires we are down to six volts and we will get 6/2 = 3 amperes of current.

In the bottom parallel connection, all the batteries are of the same voltage and connected in the same polarity, but instead of adding voltages, the voltage across the connected battery terminals is only 2 volts. In this connection arrangement, there are more batteries sustaining those 2 volts so they will last longer. So put a 2 ohm resistor across the terminals you get (2 Volts / 2 Ohms) and you get 1 ampere of current. Take one of the batteries out and you still have 2 volts over the 2 ohms and you still get 1 ampere.

So it is in the parallel connection arrangement where you can remove all but one of the batteries and still have the same current.

Why? Because in the parallel connection, removing batteries still keeps 2 volts across the batteries as long as you have at least one battery and it retains enough of its energy to provide the 2 volts.

Sincere regards,
Mike Stewart

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