Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` is

Options

A. `r_(1)-r_(2)`
B. `(r_(1)+r_(2))/(2)`
C. `(r_(1)-r_(2))/(2)`
D. `r_(1)+r_(2)`

Answer

A

Explanation

Net resistance of the circuit `=r_(1)+r_(2)+R`
net emf in series `=E+E=2E`
`i=("Net emf")/("Net resistance")`
`rArr i=(2E)/(r_(1)+r_(2)+R) " " `…(i)
It is given that as circuit is closed, potential difference across the first cell is zero. That is,
`V=E-ir_(1)=0`
`rArr i=(E)/(r_(1)) " " ` ...(ii)
Equating Eqs. (i) and (ii) , we get
`(E)/(r_(1))=(2E)/(r_(1)+r_(2)+R)`
`rArr 2r_(1)=r_(1)+r_(2)+R`
`therefore R="External resistance"=r_(1)-r_(2)`