# Why should a battery not give current in order to measure EMF?

Emf is the "potential difference (PD) across the terminals of a battery when it is giving no current to the circuit."

What does "when it is giving no current mean"? Will the PD across the terminals of battery when it is not giving current to the circuit equal to the case when the battery is giving current to the circuit?

If the internal resistance is coming into play to answer this question then I want to ask that "Internal resistance is present is present in both the cases, either the battery is giving no current or giving it to the circuit, then why do we not call the PD across battery terminals "emf" when it is giving current."

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## 1 Answer

It's hard to sure without the context, but I'd guess that the definition is given that way because all batteries have a non-zero internal resistance, $R_i$, so if a current $I$ is flowing the measured voltage is $E - IR_i$ where $E$ is the EMF of the battery. The measured voltage only equals $E$ when $I$ is zero i.e. no current is flowing.

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To produce PD across the terminals of a battery, charge will have to go from negative to positive terminal of the battery. So, shouldn't the internal resistance come into play then? –  Rafique May 22 '13 at 9:26
The internal resistance is always present, but it only causes the apparent voltage of the battery to be reduced when a current is flowing. In practice the voltmeter you use will draw some current, but voltmeters are designed to have a very high resistance so they draw very little current. –  John Rennie May 22 '13 at 9:29
So, what actually happens to the PD across battery terminals when current is supplied to the circuit and when current is not supplied to the circuit. Will the PD across battery terminals change? If it does, then why? –  Rafique May 22 '13 at 9:41
Have a look at the Wikipedia article on internal resistance I linked, and comment here if you need anything in the article clarified. –  John Rennie May 22 '13 at 10:18
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