Solution:

We are tasked to find the external resistance

$R$

when the potential difference across the first cell is zero. Let the emf of each cell be

$E$

, the internal resistances of the two cells be

$r_1$

and

$r_2$

, and the external resistance be

$R$

.

Key points:

1. Current in the circuit: The total resistance in the circuit is
   $r_1 + r_2 + R$ 

   . The current $I$ 

   is given by: 

   $$I = \frac{E + E}{r_1 + r_2 + R} = \frac{2E}{r_1 + r_2 + R}.$$ 

2. Potential difference across the first cell: The potential difference across the first cell is: 

   $$V_1 = E - I r_1.$$Since the potential difference across the first cell is zero, we set

   $V_1 = 0$:

    

   $$0 = E - I r_1.$$Substitute

   $I = \frac{2E}{r_1 + r_2 + R}$:

    

   $$0 = E - \frac{2E}{r_1 + r_2 + R} \cdot r_1.$$ 

3. Simplify the equation: Rearrange: 

   $$E = \frac{2E r_1}{r_1 + r_2 + R}.$$Divide through by

   $E$(since

   $E \neq 0$):

    

   $$1 = \frac{2r_1}{r_1 + r_2 + R}.$$Multiply both sides by

   $r_1 + r_2 + R$:

    

   $$r_1 + r_2 + R = 2r_1.$$Simplify:

    

   $$R = 2r_1 - r_1 - r_2.$$ 

   $$R = r_1 - r_2.$$ 

Final Answer:

The value of

$R$

is:

$$\boxed{R = r_1 - r_2}.$$