Two batteries of different emf and internal resistances connected in series with each other and with an external load resistor. The current reversed, the current becomes 1.0 A. the ratio of the emf of the two batteries is

Options

A. `2.5:1`
B. `2:1`
C. `3:2`
D. `1:1`

Explanation

To solve the problem, we need to analyze the situation with two batteries connected in series, each with different electromotive forces (emf) and internal resistances. We will derive the ratio of the emfs of the two batteries based on the information given about the currents.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( e_1 \) be the emf of the first battery.
- Let \( e_2 \) be the emf of the second battery.
- Let \( r_1 \) be the internal resistance of the first battery.
- Let \( r_2 \) be the internal resistance of the second battery.
- Let \( R \) be the external load resistance.

2. **Current in the Forward Direction**:
- When the current is flowing in the forward direction, the total current \( I_1 \) can be expressed as:
\[ I_1 = \frac{e_1 + e_2}{r_1 + r_2 + R} \]
- According to the problem, \( I_1 = 3 \, \text{A} \). Thus, we have:
\[ 3 = \frac{e_1 + e_2}{r_1 + r_2 + R} \quad \text{(Equation 1)} \]

3. **Current in the Reversed Direction**:
- When the current is reversed, the total current \( I_2 \) can be expressed as:
\[ I_2 = \frac{e_1 - e_2}{r_1 + r_2 + R} \]
- According to the problem, \( I_2 = 1 \, \text{A} \). Thus, we have:
\[ 1 = \frac{e_1 - e_2}{r_1 + r_2 + R} \quad \text{(Equation 2)} \]

4. **Forming the Ratio**:
- Now, we will divide Equation 1 by Equation 2 to eliminate the common denominator:
\[ \frac{3}{1} = \frac{e_1 + e_2}{e_1 - e_2} \]
- This simplifies to:
\[ 3(e_1 - e_2) = e_1 + e_2 \]
- Rearranging gives:
\[ 3e_1 - 3e_2 = e_1 + e_2 \]
- Combining like terms leads to:
\[ 3e_1 - e_1 = 3e_2 + e_2 \]
\[ 2e_1 = 4e_2 \]

5. **Finding the Ratio**:
- Rearranging gives:
\[ \frac{e_1}{e_2} = \frac{4}{2} = 2 \]

### Conclusion:
The ratio of the emf of the two batteries is:
\[ \frac{e_1}{e_2} = 2 \]