The two batteries of emf `E_(1)` and `E_(2)` having internal resistances `r_(1)` and `r_(2)` respectively are connected in series to an external resistor R. Both the batteries are gatting discharged. The above described combination of these two batteries has to produce a weaker current then when any one of the battery is connected to same resistor. For this requirement to be fulfilled.

Options

A. `(E_(2))/(E_(1))` must not lie between `(r_(2))/(r_(1)+R) and (r_(2)+R)/(r_(1))`
B. `(E_(2))/(E_(1))` must lie between `(r_(2))/(r_(1)+R) and (r_(2)+R)/(r_(1))`
C. `(E_(2))/(E_(1))` must lie between `(r_(2))/(r_(1)+R) and (r_(1))/(r_(2)+R)`
D. `(E_(2))/(E_(1))` must not lie between `(r_(2))/(r_(1)+R) and (r_(1))/(r_(2)+R)`

Answer

To solve the problem, we need to analyze the situation where two batteries with different EMFs (E1 and E2) and internal resistances (r1 and r2) are connected in series to an external resistor R. We want to find the condition under which the current produced by this combination is weaker than the current produced by either battery when connected individually to the same resistor R.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- When two batteries are connected in series, the total EMF (E_total) is the sum of the individual EMFs:
\[ E_{total} = E_1 + E_2 \]
- The total internal resistance (R_internal) is the sum of the individual internal resistances:
\[ R_{internal} = r_1 + r_2 \]

2. **Applying Ohm's Law**:
- The total resistance in the circuit when both batteries are connected in series to the external resistor R is:
\[ R_{total} = R + R_{internal} = R + r_1 + r_2 \]
- The current (I_series) through the circuit can be calculated using Ohm's Law:
\[ I_{series} = \frac{E_{total}}{R_{total}} = \frac{E_1 + E_2}{R + r_1 + r_2} \]

3. **Current with Individual Batteries**:
- When only battery 1 is connected to the resistor R, the current (I1) is:
\[ I_1 = \frac{E_1}{R + r_1} \]
- When only battery 2 is connected to the resistor R, the current (I2) is:
\[ I_2 = \frac{E_2}{R + r_2} \]

4. **Condition for Weaker Current**:
- For the combination of the two batteries to produce a weaker current than either battery alone, we need:
\[ I_{series} < I_1 \quad \text{and} \quad I_{series} < I_2 \]
- This gives us two inequalities:
\[ \frac{E_1 + E_2}{R + r_1 + r_2} < \frac{E_1}{R + r_1} \]
\[ \frac{E_1 + E_2}{R + r_1 + r_2} < \frac{E_2}{R + r_2} \]

5. **Simplifying the Inequalities**:
- Rearranging the first inequality:
\[ (E_1 + E_2)(R + r_1) < E_1(R + r_1 + r_2) \]
This simplifies to:
\[ E_2(R + r_1) < E_1 r_2 \]
- Rearranging the second inequality:
\[ (E_1 + E_2)(R + r_2) < E_2(R + r_1 + r_2) \]
This simplifies to:
\[ E_1(R + r_2) < E_2 r_1 \]

6. **Conclusion**:
- The conditions derived from the inequalities provide the relationship between the EMFs and internal resistances of the batteries that must be satisfied for the combined current to be weaker than that of either battery alone.