Two cells of internal resistance `r_1 and r_2` and of same emf are connected in series across a resistor of resistance R. If the terminal potential difference across the call of internal resistance `r_1` is zero , then the value of R is

Options

A. `R=2(r_1+r_2)`
B. `R=r_2-r_1`
C. `R=r_1-r_2`
D. `R=2(r_1-r_2)`

Answer

C

Explanation

Let the emf of each cell be `epsilon` .
The cells are connected in series across R as shown in figure .
Since the cells are in series , so their equivalent emf is

`epsilon_(ep)=epsilon+epsilon=2epsilon`
and their equivalent internal resistance is
`r_(eq)=r_1+r_2`
The current in the circuit is
`I=(epsilon_(eq))/(r_(eq)+R)=(2epsilon)/(r_1+r_2+R)" "...(i)`
The terminal potential difference across the cell of internal resistance `r_1` is
`V_1 = epsilon=Ir_1`
But as per question `V_1=0`
`:. 0 = epsilon- Ir_1 " or " I=epsilon/r_1 " "...(ii)`
Equating (i) and (ii) , we get
`(2epsilon)/(r_1+r_2+R)=epsilon/r_1" or "2r_1=r_1+r_2+R`
or, `R=2r_!-r_1-r_2=r_1-r_2`