A cell of emf 2 V and internal resistance `1.2Omega` is connected with an ammeter of resistance `0.8Omega` and two resistors of `4.5Omegaand9Omega` as shown in the diagram below:

What is the potential difference across the terminals of the cell ?

Explanation

The resistance of `4.5Omegaand9Omega` are connected in parallel.
`therefore` Equivalent resistance,
`R_(1)=((4.5xx9))/((4.5+9))=(40.5)/(13.5)=3Omega`
Total resistance in the circuit (R )
= `1.2+0.8+3=5Omega`
Potential difference across the terminals of the cell (V) = Total p.d. in the external circuit
= E - Ir
= 2 - `(0.4xx1.2)`
= 2 - 0.48 = 1.52 volts