To the parallel combination of two resistances `3 Omega` and `1 Omega`, a series combination of resistances `2.15 Omega` and `1 Omega` and a battery are connected. The interval resistance of the battery is `0.1 Omega` and the emf is 2V. Determine the values of current flowing through the resistances. Draw the diagram of the circuit.

Explanation

Refer to fig.1.17 for the circuit diagram the equivalent resistance of the parallel combination of `3 Omega and 1 Omega` is
`R=(3times1)/(3+1)=3/4=0.75 Omega`

This equivalent resistance is in series with the other resistances So, the main current of the ciruit is
`l=2/(0.75+2.15+1+0.1)=2/4=0.5A`
This main current flows through the resistances `2.15 Omega` and `1 Omega` connected in series.
Now current throught `3 Omega`
`I_1=Itimes1/(3+1)=0.5times1/4=0.125A`
and current through `1 Omega`,
`I_2=I-I_1=0.5-0.125=0.375A`