In the circuit shown below, `E_(1)=17V,E_(2)=21V,R_(1)=2Omega,R_(2)=3Omega and R_(3)=5Omega`. Using Kirchhoff.s laws, find the currents flowing through the resistors `R_(1),R_(2)` and `R_(3)`. (Internal resistance of each of the batteries is negligible.)

Explanation

Given : `E_(1)=17V,E_(2)=21V,R_(1)=2Omega,R_(2)=3Omega` and `R_(3)=5Omega`
In loop AFEB,
`I_(1)R_(1)+(I_(1)+I_(2))R_(3)=E_(1)`
`I_(1)xx2+(I_(1)+I_(2))5=17`
`2I_(1)+5I_(1)+5I_(2)=17`
`7I_(1)+5I_(2)=17" "........(1)`
In loop BCDE,
`I_(1)R_(2)+(I_(1)+I_(2))R_(3)=E_(3)`
`I_(2)xx3+(I_(1)+I_(2))5=21`
`5I_(1)+8I_(2)=21" "......(2)`
On multiplying equation (1) by 5 and equation (2) by 7, we get
`35I_(1)+25I_(2)=85`
`35I_(1)+56I_(2)=147`
`(---)/(-31I_(2)=-62)`
`I_(2)=2A`
`7I_(1)+5I_(2)=17` [from equation (1)]
`7I_(1)+10=17`
`7I_(1)=7`
`I_(1)=1A`
`I=I_(1)+I_(2)`
`=1+2=3A`