Two cells of same emf `epsilon` but of different internal resistances `r_1` and `r_2` are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is

Options

A. `r_1+r_2`
B. `r_1-r_2`
C. `r_2-r_1`
D. `r_1^2//r_2`

Answer

B

Explanation

As both both cells are in series , the circuit current
`I=(epsilon+epsilon)/(r_1+r_2+R)=(2epsilon)/(r_1+r_2+R)`
As terminal potential drop across `1^(st)` cell is zero , hence
`V_1=epsilon-Ir_1=epsilon-(2epsilon)/((r_1+r_2+r))r_1=0`
`rArr epsilon=(2epsilonr_1)/((r_1+r_2+R))` or `r_1+r_2+R=2r_1` or `R=(r_1-r_2)`