Question

A cell of emf 1.8v is connected to an external resistance of 2ohms .when potential difference recorded at the ends of resistance is 1.6v . calculate internal resistance of the cell.

Answer 1

all the resistance of battery, ammeter, and resistance of wire are in series combination.

So overall equivalent resistance = 2 +4.5 + 0.7 ohm = 7.2 ohm

As resistance of battery = 2 ohm , Resistance of wire = 4.5 ohm , Resistance of ammeter = 0.7 ohm.

Now according to Ohms law we know

Current(I) = voltage(V)/ resistance(R)

So current in the ammeter = 1.8/7.2 = 0.25 ampere. This is the answer of question (A)

Here potential difference (V) = current(I)* resistance(R) = 0.25* 5.2 = 1.3 V

As wire resistance and ammeter resistances are in the series combination.

So 1.3V is the potential difference of the circuit .