Question

Define electromotive force of a cell.

Hint:

Remember the measurement rule: {Open-circuit} voltmeter reading at the terminals gives emf (\(I=0 ⇒ V_{\text{terminal}}=\mathcal{E}\)). Under load, use \(V_{\text{terminal}}=\mathcal{E}-Ir\).

Answer 1

Step 1: Identify the source of the push. Inside a cell a non-electrostatic (chemical) action separates charge and maintains a steady potential difference between its plates. The measure of this driving action is the electromotive force.

Step 2: Give the definition as maximum potential difference. The EMF of a cell is the maximum potential difference between its two terminals, which appears when the cell is not delivering any current (open-circuit condition). At that instant no current flows, so there is no drop across the internal resistance and the full driving voltage is available.

Step 3: Restate energetically and note units. In energy terms it is the total energy the cell supplies to each coulomb of charge that passes through the whole circuit, \(\varepsilon = W/q\), measured in volts.
\[\boxed{\varepsilon = \text{open-circuit terminal p.d.} = \dfrac{W}{q}}\]