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

(a) Currentin the circuit is given by Ohm's law.
Net resistance of the circuit `= r_(1) + r_(2) + R`
Net emf in series `= E + E = 2E`
Therefore, form Ohm's law, current in the circuit
`i= ("Net emf")/("Net resistance")`
`implies i= (2E)/(r_(1) + r_(2) + R)`
It is given that, as circuit is closed, potential difference across the first cell is zero. That is,
`V = E = ir_(1) = 0`
`implies i= (E)/(r_(1))`
Equating Eqs. (i) and (ii), we get `(E)/(r_(1)) = (2E)/(r_(1) + r_(2) + R)`
`2r_(1) = r_(1) + r_(2) + R`
`R = ` external resistance `= r_(1) - r_(2)`