Two cells of emf $$2E$$ and $$E$$ with internal resistance $$r_1$$ and $$r_2$$ respectively are connected in series to an external resistor $$R$$ (see figure). The value of $$R$$, at which the potential difference across the terminals of the first cell becomes zero is:

We need to determine the value of the external resistance $$R$$ at which the potential difference across the terminals of the first cell becomes zero.

1. Identify the Circuit Parameters

From the circuit diagram, two cells are connected in series aiding each other (their polarities are aligned in the same direction to drive current together through the external resistor $$R$$):

• First Cell: Electromotive force = $$2E$$, Internal resistance = $$r_1$$
• Second Cell: Electromotive force = $$E$$, Internal resistance = $$r_2$$

2. Calculate the Total Current ($$I$$) in the Circuit

Using Ohm's law for the complete series loop, the total current flowing through the circuit is equal to the total net electromotive force divided by the total equivalent resistance:

$$I = \frac{\text{Net EMF}}{\text{Total Resistance}} = \frac{2E + E}{R + r_1 + r_2}$$

$$I = \frac{3E}{R + r_1 + r_2}$$

3. Set Up the Condition for Zero Potential Difference

The terminal potential difference ($$V_1$$) across the first cell during discharge is given by the expression:

$$V_1 = 2E - I r_1$$

According to the problem statement, this terminal potential difference is equal to zero ($$V_1 = 0$$):

$$2E - I r_1 = 0 \implies 2E = I r_1$$

4. Solve for the External Resistance ($$R$$)

Substitute our expression for total current ($$I$$) from Step 2 into the condition equation:

$$2E = \left( \frac{3E}{R + r_1 + r_2} \right) r_1$$

Divide both sides of the equation by $$E$$ to eliminate it:

$$2 = \frac{3r_1}{R + r_1 + r_2}$$

Cross-multiply to clear the denominator:

$$2(R + r_1 + r_2) = 3r_1$$

$$2R + 2r_1 + 2r_2 = 3r_1$$

Isolate the term containing $$2R$$ by shifting the internal resistance terms to the right-hand side:

$$2R = 3r_1 - 2r_1 - 2r_2$$

$$2R = r_1 - 2r_2$$

$$R = \frac{r_1}{2} - r_2$$

Conclusion

The value of the external resistor $$R$$ at which the terminal voltage of the first cell drops to zero is $$\frac{r_1}{2} - r_2$$, which corresponds to Option B.