Question

Two cells, having the same e.m.f are connected in series through an external resistance R. Cells have internal resistance $r_1$ and $r_2$ ( $r_1$ > $r_2$) respectively. When the circuit is closed, the potential difference across the first cell is zero. the value of R is

• A. $\frac{r_1+r_2}{2}$
• B. $\frac{r_1-r_2}{2}$
• C. $r_1+r_2$
• D. $r_1-r_2$

Answer 1

The correct option is D $r_1-r_2$

Two cells of emf E are connected in series through external resistance R.

Thus, total resistance is R + r1 + r2

Thus, current across the circuit is total voltage divided by total resistance.

I=E+E/R+r1+r2

I=2E/R+r1+r2

Now, when circuit is closed, the voltage across first cell is zero. Thus, we have

Ir1=E

E/r1=2E/R+r1+r2

R=r1-r2

This is the value of R in terms of r1 and r2.