The emf of a cell is the potential difference across its terminals when it is in

Answer Details

The electromotive force of a cell is the total energy the cell gives to each unit of charge driven round the whole circuit, including the energy wasted inside the cell itself. Every real cell has internal resistance \(r\), so once a current \(I\) flows there is a voltage drop \(Ir\) inside the cell and the reading at the terminals is only \[V = E - Ir.\] The terminal potential difference therefore equals the emf exactly when \(I = 0\), and the current is zero when the circuit is broken. Hence the emf is the terminal potential difference measured on open circuit.

In practice this is why a voltmeter of very high resistance, or a potentiometer used at balance, gives a true emf reading: both draw negligible current, so the \(Ir\) term vanishes. Connect the same cell to a lamp and the terminal reading falls, which is the everyday observation that an old battery reads its full value until it is asked to supply current.

Choosing a closed circuit is the common error: in a closed circuit the reading is the terminal potential difference, which is always less than the emf by \(Ir\) for a discharging cell. Series and parallel refer to how cells or components are combined, not to the condition under which emf is defined, so neither answers the question. In an examination, use two readings to separate the two quantities: an open-circuit reading gives \(E\), and a loaded reading with known \(I\) then gives \(r = \dfrac{E - V}{I}\).