Question

Explain why is the p.d across the terminals of a cell more in an open circuit and reduced in a closed circuit.

Answer 1

n ideal battery or cell has no internal resistance$,$ $i.e.,$ $r=0.$

So there is no voltage drop across the materials in side the battery$,$ When it is connected in a circuit with some external elements$.$

When a battery is not connected in a circuit or it is open$,$ then there is no current passing through it$.$ So the emf the potential difference observed by measurement$.$

When a battery has an internal resistance $r,$ and it is connected through a circuit$,$ there is a current $I$ flowing in the circuit$.$ Hence$,$ there is a voltage drop across this internal resistance$.$

So $P.d.$ across battery$=emf-r.I$