Half Batteries Not Half Current
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.
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.
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-
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.
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
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
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.
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Update: June 2012