A battery of internal resistance 2 Ohms is connected to an external resistance of 10 ohms. The current is .5 A. What is the emf of the battery?

The correct answer is 6.0 V. When I calculated it like a series circuit with 2 resistors I got 6 V for the emf. But I thought the emf did not include the internal resistance, so I calculated it without the internal resistance and got 5.0 V. I was wondering if someone could clarify this for me?enter image description here


When current flows, inside the source there is always some voltage drop, so the voltage on the terminals is lower than in the static situation.

We model this behaviour by simple model depicted on your picture; we assume that the real source consists of idealized source -| |- that gives the emf and has not resistance, and of ordinary ohmic resistance (2 $\Omega$ in your case).

By emf of the source we mean the number giving voltage due to the idealized source -||- rather than the total voltage on the real source when current flows, because the latter voltage depends on the current and that depends on what elements are connected to the circuit.

When there is no current, there is no voltage drop on the internal resistance, so we can say the emf of the source is equal in magnitude to the voltage the source creates on its outer terminals when current does not flow.


If $E$ is the emf (open-circuit voltage) of the battery and $r_s$ is the internal resistance, the equation relating the series current $I_S$ through an external resistance $R_L$ is given by:

$$I_S = \frac{E}{r_s + R_L}$$

Now, the voltage across the battery terminals $V_{BAT}$ is given by

$$V_{BAT} = E - I_S \cdot r_s = E\frac{R_L}{r_s + R_L}$$

So, even though we can't 'see' the internal resistance and measure the voltage across it, we can infer its existence by the fact that the terminal voltage depends on the current or, equivalently, the external resistor.

Of course, there is no actual resistor inside the battery but, for many applications, modelling this terminal voltage dependence on current with a resistor is adequate.


It's much easier to use the EMF equation to work this out:

$${\rm EMF} = I(R+r)$$

Where EMF is what you are calculating (in volts)

$I = \rm Current$

$R =$ Resistance of the component (in this case, the resistor)

$r =$ Internal Resistance

Plugging in the numbers, you get this:

$${\rm EMF} = 0.5(10+2)$$ $${\rm EMF} = 0.5(12)$$ $${\rm EMF} = 0.5 \times 12$$ $${\rm EMF} = 6~\rm V$$

Hope this helped!

  • $\begingroup$ In the units currents is in amperes and resistance in ohms. $\endgroup$ – SAKhan May 6 '18 at 0:42

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