Two cells , one of emf 1.2 V and internal resistance `0.5Omega` the other of emf 2 V and internal resiatnce `0.1Omega` are connnected in series with an external resistance of `5Omega` . What is the current through this resistor ?

Explanation

Applying Kirchhoff's first law to the point A we have ,
`I_(1)+I_(2)-I=0 " or , " I_(2)=I-I_(1)" "... (1) `
Applying Kirchhoff's seond law to the loop ACBDA we have ,
`-I_(1)xx0.5+I_(2)xx0.-1.2+2`
`"or , "-I_(1)xx0.5+(I-I_(1))xx0.1 = 0.8`
`"or, " 0.1I-0.6I_(1)=0.8` ...... (2)
For the loop ADBFA
`-I_(2)xx0.1-Ixx5=-2`
`"or, "(I-I_(1))xx0.1+5l=2"or, " 5.1I_(1)=2` ...... (3)

Now , multiplying equation (3) by 6 subtracting equation (2) from it we have ,
`30.5I=11.2 " or, "I=(11.2)/(30.5)=0367A`