A 5V battery with internal resistance `2Omega` and a 2V battery with internal resistance`1Omega` are connected to a `10Omega` resistor as shown in the figure.

The current in the `10Omega` resistor is

Options

A. 0.03 A from `P_(1) " to " P_(2)`
B. 0.03 A from `P_(2) " to " P_(1)`
C. 0.28 A from `P_(1) " to " P_(2)`
D. 0.28 A from `P_(2) " to " P_(1)`

Answer

D

Explanation

The current distribution in the various parts is as shown in the figure.

Applying Kirchhoff's law to the loop `AP_(2)P_(1)DA`,
we get, `10 I_(1)+2 (I_(1)+I_(2))-5=0`
`therefore 12 I_(1)+ 2 I_(2)=5`
and for the loop, `P_(2) BCP_(1) P_(2)`,
`2+I_(2)xx 1-10 I_(1)=0`
`therefore I_(2)- 10 I_(1)= -2`
or `10 I_(1)-I_(2)=2`
`therefore 20 I_(1)-2 I_(2)=4`
Add (1) and (3) `therefore 32 I_(1)=9`
`therefore I_(1)=(9)/(32)=0.28` A from `P_(2) " to " P_(1)`.