Two batteries of e.m.f. `4V` and `8V` with internal resistances `1 Omega` and `2 Omega` are connected in a circuit with a resistance of `9 Omega` as shown in figure. The current and potential difference between the points `P` and `Q`

Options

A. `(1)/(2)A,9 V`
B. `(1)/(12)A, 12 V`
C. `(1)/(3)A, 3 V`
D. `(1)/(6)A,4 V`

Answer

c

Explanation

Net emf `=E_(2)-E_(1)= 8V-4V=4V`
Net resistance `=R+r_(1)+r_(2)`
`=9+1+2`
`12 Omega`
From Ohm's law
`V= iR`
`i=(V)/(R )=(4)/(12)=(1)/(3)A`
Potential difference between `P` and `Q`
`=` Potential difference across `R`
`=IR=(1)/(3)xx9= 3V`