Internal Resistance

In one sentence: A real battery has a small resistance inside it, so its output voltage drops when it pushes a current through a load.

$V = \varepsilon - I r$

Read it out loud: “V equals epsilon minus I times r. Read V as the terminal voltage, the voltage across the battery's output terminals while it supplies current. Epsilon is the Greek letter epsilon, said EP-sih-lon, and it is the electromotive force, or emf, the battery's voltage with no current flowing. I is the current the battery supplies, in amperes. r is the internal resistance, the resistance inside the battery itself.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Volts (V) and voltage: Voltage is the push that drives electric current, a bit like the pressure in a water pipe. Its unit is the volt, written V. A 9 V battery gives a stronger push than a 1.5 V cell. One volt means one joule of energy is given to each coulomb of charge that passes through the source.
Current, the ampere (A) and the ohm ( $\Omega$ ): Current is the flow of electric charge, and its unit is the ampere, written A. One ampere is one coulomb of charge per second. Resistance is how hard a part makes it for current to pass, and its unit is the ohm, written $\Omega$ . Ohm’s law says voltage equals current times resistance, $V = I R$ .
Power, the watt (W) and $I^2 R$: Power is the rate at which energy is used or made. Its unit is the watt, written W, and one watt is one joule per second. A current $I$ through a resistance $R$ turns $I^2 R$ of its energy into heat each second. Here $I^2$ means $I$ times $I$ , so doubling the current makes the power four times bigger.
Terminals and the terminal voltage: The terminals are the two metal ends where a battery connects to a circuit, the plus and minus poles. The terminal voltage is the voltage measured across those two ends while the battery is supplying current. It is usually a little lower than the battery’s rated voltage, and the difference is the subject of this page.
The Greek letter epsilon ( $\varepsilon$ ) and emf: The Greek letter $\varepsilon$ , said EP-sih-lon, is the usual symbol for the electromotive force, or emf. Despite its name, emf is a voltage, not a force. It is the voltage of the battery when no current is flowing. It comes from the chemistry inside the cell, and it is measured in volts.
A short circuit: A short circuit is a connection with almost no resistance between the two ends of a source, so the current is limited only by the source’s own internal resistance. A short circuit can make a battery very hot, so it is dangerous. The current in a short circuit is the emf divided by the internal resistance.

What each symbol means

Symbol	Meaning
$V$	The terminal voltage, the voltage across the battery’s output terminals while it supplies current, measured in volts (V). It is what a meter reads across the battery during use. A healthy 9 V battery feeding a small load still reads close to 9 V, a little lower than its rating.
$\varepsilon$	The electromotive force (emf), the battery’s voltage with no current flowing, measured in volts (V). A fresh alkaline AA cell has an emf of about 1.5 V. The emf comes from the chemistry inside the battery, so it is set by the cell’s materials.
$I$	The current that the battery supplies to the circuit, measured in amperes (A). A small radio may draw a few hundredths of an ampere, while a car starter motor can draw hundreds of amperes for a few seconds.
$r$	The internal resistance of the battery, the resistance of the materials inside the cell, measured in ohms ( $\Omega$ ). It is usually small, often a fraction of an ohm for a good battery, and it grows as the battery ages.

The formula piece by piece

$\varepsilon$: Start with the emf, the voltage the battery would deliver if no current flowed. It is the starting push, measured in volts. Everything that follows is a loss subtracted from this number.
$I r$: The voltage that is used up inside the battery. The current $I$ flows through the internal resistance $r$ , and by Ohm’s law the voltage lost equals $I$ times $r$ , in volts. The more current the battery supplies, the more voltage it loses inside itself.
$\varepsilon - I r$: The emf minus the voltage lost inside the battery. What is left is the voltage available at the terminals, the voltage a meter reads across the battery while it is working.
$V = \varepsilon - I r$: The terminal voltage equals the emf minus the current multiplied by the internal resistance, so the voltage at the battery’s terminals falls as the load draws more current.

What the formula is doing, and why

An ideal battery would hold its emf, $\varepsilon$ , however much current you draw. A real battery is not ideal. It has a small resistance $r$ inside it, made from the electrolyte, which is the chemical paste or liquid that carries charge inside the cell, and from the electrodes and connections. This resistance is called the internal resistance. It sits in series with the circuit outside the battery, so the same current flows through both.

Why subtract? When current $I$ flows out of the battery, it also flows through $r$ . By Ohm’s law, that causes a voltage drop of $I r$ inside the battery. The voltage left at the terminals is therefore $V = \varepsilon - I r$ . The subtraction is the voltage spent pushing the current through the battery’s own resistance. More current means a bigger drop, so the terminal voltage falls.

Why it matters when you draw current. With no current flowing, the drop is zero, so the terminal voltage equals the emf. That is why a meter across a battery that is not connected to anything reads close to its full rating. A meter draws almost no current, so it shows almost the emf. Under a heavy load, the same battery reads noticeably lower, because the drop $I r$ has grown.

What emf really is, and where the heat goes. The emf is the energy the source gives to each coulomb of charge, measured in volts. Internal resistance grows as a battery ages, so an old battery sags more under the same load. The current through $r$ also turns $I^2 r$ watts of power into heat inside the cell, which is why a battery can feel warm when it delivers a large current.

When the simple model is not enough. The formula treats the battery as a constant emf in series with one fixed resistance. Real batteries can change as they discharge, with the emf and the internal resistance both shifting over time. So the formula is best used for a short time, over which those changes are small.

An everyday picture

Think of a tank of water with a pump that pushes water out through a pipe. The pump gives the water its push, which is the emf. When the water flows fast, some of that push is used up forcing the water through a narrow outlet inside the tank. The pressure at the tap is lower than the pump’s full push, and the faster the flow, the bigger the loss. In the battery, the narrow outlet is the internal resistance $r$ , and the water flow is the current $I$ .

Worked example: A 9 V battery with internal resistance driving a load

A 9 V battery, like the one in a small radio, has an emf of 9 V and an internal resistance of $0.5\ \Omega$ . It drives a load resistor of $4.5\ \Omega$ . Find the current, the terminal voltage, and the power that turns into heat inside the battery. The numbers are chosen for practice.

Find the current for the whole circuit. The battery and the load are in series, so the same current flows through the internal resistance and the load. The two resistances add: $R + r = 4.5\ \Omega + 0.5\ \Omega = 5\ \Omega$ . Ohm’s law for the whole loop gives $I = \frac{\varepsilon}{R + r} = \frac{9\ \text{V}}{5\ \Omega} = 1.8\ \text{A}$ . So the battery sends 1.8 coulombs of charge around the loop each second.
Find the terminal voltage with the formula. The terminal voltage is the emf minus the drop inside the battery. The drop is $I r = 1.8\ \text{A} \times 0.5\ \Omega = 0.9\ \text{V}$ , the current times the internal resistance. So $V = \varepsilon - I r = 9\ \text{V} - 0.9\ \text{V} = 8.1\ \text{V}$ . The battery’s output is 0.9 volts below its rating, because of the drop inside it.
Check the answer from the load side. The load sits directly across the battery’s terminals, so the voltage across the load must equal the terminal voltage. Ohm’s law for the load gives $V = I R = 1.8\ \text{A} \times 4.5\ \Omega = 8.1\ \text{V}$ . This matches the 8.1 V from the first route, so the two sides of the circuit agree.
Find the power lost inside the battery. Power is the rate of using energy, in watts. In the internal resistance it is $P = I^2 r$ , where $I^2$ means $I$ times $I$ . So $P = (1.8\ \text{A})^2 \times 0.5\ \Omega = 3.24\ \text{A}^2 \times 0.5\ \Omega = 1.62\ \text{W}$ . The units check out, because amperes squared times ohms gives watts. This 1.62 watts becomes heat inside the cell, not useful work.
Check the energy budget. The emf delivers $\varepsilon I = 9\ \text{V} \times 1.8\ \text{A} = 16.2\ \text{W}$ in total. The load takes $I^2 R = 3.24 \times 4.5 = 14.58\ \text{W}$ , and the internal resistance takes 1.62 W. The two parts add to $14.58 + 1.62 = 16.2$ W, the same as the total. Energy is not created or lost, only shared between the load and the battery’s inside.

Answer: The battery supplies 1.8 A and its terminal voltage falls to 8.1 V, which is 0.9 V below its 9 V emf. About 1.62 W of power is lost as heat inside the battery itself, and the load receives the other 14.58 W.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Using the emf as the voltage you measure under load. Under load the terminal voltage is lower by $I r$ , so a meter across a loaded battery reads less than the emf.
Forgetting that the internal resistance is in series with the load. Both resistances carry the same current, so add them together to get the total resistance in Ohm’s law.
Thinking a short circuit draws an infinite current. A real battery is limited by $r$ , so the short-circuit current is $\varepsilon / r$ . For the 9 V battery that is $9\ \text{V} / 0.5\ \Omega = 18\ \text{A}$ , which is large but finite.
Assuming a battery’s label voltage is always what it delivers. Under heavy load the output can sag well below the label, especially for an old battery with a large internal resistance.

Where you will meet it

Testing a battery’s health by measuring its voltage under a known load. A large sag points to a high internal resistance.
Car starter batteries, which sag noticeably while the starter motor draws a large current for a few seconds.
Designing power supplies and battery packs, where the output resistance sets how much the voltage drops as the load grows.
Estimating the heating inside a battery that delivers a high current. The heat in watts is $I^2 r$ , and it grows with the square of the current.

Questions people ask

Why does my battery read a full voltage on a meter but drop when it runs a device?

The meter draws almost no current, so there is almost no drop across the internal resistance, and the reading is close to the emf. When the device draws a real current, the drop $I r$ pulls the terminal voltage down.

Is internal resistance the same as a resistor in the circuit?

It is a resistance, so it follows Ohm’s law in the same way, but it sits inside the battery. You cannot remove it, and it always adds to the total resistance of the loop.

What is emf?

The electromotive force is the energy the source gives to each coulomb of charge, measured in volts. It equals the battery’s voltage when no current is flowing. The name is old, and the quantity is a voltage, not a force.

Does internal resistance make a battery hot?

Yes, a little. The current through $r$ turns $I^2 r$ watts into heat inside the cell. The heating is greater at high currents, because the power grows with the square of the current.

Ohm's Law: Ohm’s law says the push on a wire, called voltage, equals the flow through it, called current, times how hard the wire resists that flow, called resistance.
Maximum Power Transfer: A load receives the most power when its resistance matches the source’s resistance, but then half the power is lost inside the source.
Electric Power: Electric power tells you how fast a device uses energy. You get it by multiplying the voltage, the push, by the current, the flow, and the answer is measured in watts.
Kirchhoff's Voltage Law: Kirchhoff’s voltage law says that if you walk all the way around a closed circuit loop, the voltage you gain equals the voltage you lose, so the net change is zero.