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Wize University Physics Textbook (Master) > DC Circuits

Real Batteries

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Real Batteries and Internal Resistance




In reality, there is an internal resistance associated to batteries.

  • The voltage provided by the battery is known as the electromotive force and it is shown as E\mathcal{E} or EMFEMF. It is the potential difference across the battery with no current present.
  • The terminal voltage VV is the actual voltage across the battery, and is usually less than the electromotive force because of the voltage drop across the internal resistance.












The terminal voltage is given by:
 V=E−Ir \boxed{\ V=\mathcal{E}-Ir \ }

where rr is the internal resistance, and II is the current passing through the battery.
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Exam Tip
To account for internal resistance, add it as a separate resistor in series with the battery.


Using Ohm's Law we get:

 I=ERnet=ER+r \boxed{ \ I=\dfrac{\mathcal{E}}{R_{net}}=\dfrac{\mathcal{E}}{R+r} \ }


where RnetR_{net} is the total resistance of the entire circuit, given by RR (all the resistors combined) plus rr (the internal resistance of the battery, in series with it).
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Example: Internal Resistance and Power


A real battery with an electromotive force of 2020 V is attached to a light bulb with resistance of 22 Ω which is dissipating energy with a power of 5050 W. Find the size of the internal resistance and the terminal voltage of the battery.


Use the power of the resistor to find the current through it, which must be the current going through the whole circuit:

P=I2R   →   I=PRP=I^2R \ \ \ \to \ \ \ I=\sqrt{\dfrac{P}{R}}


This the current can be written as:

I=ERnet=ER+rI=\dfrac{\mathcal{E}}{R_{net}}=\dfrac{\mathcal{E}}{R+r}


Solving for the internal resistance we get:

R+r=EIR+r=\dfrac{\mathcal{E}}{I}

r=EI−Rr=\dfrac{\mathcal{E}}{I} -R


Substitute the current from above to get:

r=ERP−Rr=\mathcal{E}\sqrt\dfrac{R}{P} -R

=20250−2=20\sqrt\dfrac{2}{50} -2

=2=2 (Ω)


The terminal voltage is:

V=E−IrV=\mathcal{E}-Ir

=E−rPR=\mathcal{E}-r\sqrt\dfrac{P}{R}

=20−2502=20-2\sqrt\dfrac{50}{2}

=10=10 (A)

Practice: Internal Resistance vs. Voltage and Current


What is the internal resistance of a voltage source if its terminal potential drops by 2.02.0 V when the current supplied increases by 5.0 5.0 A?