Two resistors of resistances `2Omega` and `4Omega` are connected in parallel. Two more resistors `3Omega` and `6Omega` are also connected in parallel. These two combinations are in series with a battery of emf 5 V and internal resistance `0.7Omega`. Calculate current through the `6Omega` resistor.

Explanation

We know that, `R_(p)=(R_(1)R_(1))/(R_(1)+R_(2))`
Hence, `R_(P_(1))=(2xx4)/6=4/3Omega`
`R_(p_(2))=(3xx6)/9=2Omega`
But, `R_(s)=R_(p_(1))+R_(p_(2))=4/3+2=10/3Omega`
R equivalent = `R_(s)+r`
where, `r=0.7=(7/10)`
`=10/3+7/10=(100+21)/(30)=121/30Omega`
`R_("eq")=4.03Omega`
We also know that, `I=E/(R_("eq"))`
hence, `I=5/(4.03)=1.24` A
and `I_(2)=(IR_(1))/(R_(1)+R_(2))=((1.24)(3))/(3+6)`
i.e., `I_(2)=0.4133` A