Now, I know that emf is measured using an ammeter, while potential difference is measured using a voltmeter. But why is that so? What do we mean when we say the emf of the battery is say, 3V, and the potential difference is 2V or anything? How do we define these terms? What is emf and the current supplied by the battery? Are they the same thing?
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What do we mean when we say the emf of the battery is say, 3V
It essentially means that the battery open-circuit voltage is 3V. That is, if you have a battery that is not connected to a load and you measure the voltage across with an ideal voltmeter, you will measure 3V
and the potential difference is 2V or anything?
A physical battery cannot supply arbitrarily large current due to what is called internal resistance. As the load on the battery is increased, the terminal voltage drops further below the open-circuit voltage. For an easy if unrealistic example, stipulate that the battery has $1\Omega$ internal resistance and that it is connected to a $2\Omega$ load.
It's easy to calculate that load current is $1A$ and that the voltage across the battery is $3V - 1A \cdot 1\Omega = 2V$
How do we define these terms? What is emf and the current supplied by the battery? Are they the same thing?
You are expected to do some basic research before asking questions here so I won't spend any time on these basic questions.
Alfred Centauri
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