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 would be the reading on the Ammeter ?

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`
Reading of the ammeter = Current in the circuit (I)
= `("Total e.m.f.(E)")/("Total resistance (R)")`
= `2/5` = 0.4 ampere