7 V battery with internal resistance `2Omega and 3V` battery with internal resistance `1Omega` are connected to a `10Omega` resistors as shown in figure, the current in `10Omega` resistor is

Options

A. 0.27A
B. 0.31A
C. 0.031A
D. 0.53

Answer

C

Explanation

Using Kirchoff's law in loop `A P_(2) P_(1) DA`
`therefore 10I_(1)+2I-7=0`
`10I_(1)+2I=7...(i)`
Using Kirchhoff's law in loop `P_(2)P_(1)CBP_(2)`
`-3+I(I-I_(1))-10 I_(1)=0`

`I-11I_(1)=3, I=3+11 I_(1) ....(ii)`
From (i) and (ii)
`10I_(1)+2(3+11I_(2))=7 10 I_(1)+6+22 I_(1)=7`
`therefore 32 I_(1)=I,I_(1)=1//32=0.031A`