When a current of 2 A flows in a battery from negative to positive terminal, the potential difference across it is 12 V. If a current of 3 A flowing in the opposite direction produces a potential difference of 15 V, the emf of the battery is

Options

A. `12.6V`
B. `13.2V`
C. `13.5V`
D. `14.0V`

Explanation

To solve the problem, we need to determine the electromotive force (emf) of the battery based on the given current and potential differences. We will use the relationship between emf, terminal voltage, current, and internal resistance.

### Step-by-Step Solution:

1. **Identify the given values:**
- For the first case:
- Current \( I_1 = 2 \, \text{A} \)
- Potential difference \( V_1 = 12 \, \text{V} \)
- For the second case:
- Current \( I_2 = 3 \, \text{A} \) (in the opposite direction)
- Potential difference \( V_2 = 15 \, \text{V} \)

2. **Write the equations for terminal voltage:**
- The terminal voltage \( V \) can be expressed as:
\[ V = E - I \cdot R \]
- For the first case:
\[ V_1 = E - I_1 \cdot R \implies 12 = E - 2R \quad \text{(Equation 1)} \]
- For the second case (current flowing in the opposite direction):
\[ V_2 = E - I_2 \cdot R \implies 15 = E - 3R \quad \text{(Equation 2)} \]

3. **Rearrange both equations to express \( E \):**
- From Equation 1:
\[ E = 12 + 2R \quad \text{(Equation 3)} \]
- From Equation 2:
\[ E = 15 + 3R \quad \text{(Equation 4)} \]

4. **Set the two expressions for \( E \) equal to each other:**
\[ 12 + 2R = 15 + 3R \]

5. **Solve for \( R \):**
- Rearranging gives:
\[ 12 - 15 = 3R - 2R \implies -3 = R \]
- Thus, \( R = 3 \, \Omega \).

6. **Substitute \( R \) back into one of the equations to find \( E \):**
- Using Equation 3:
\[ E = 12 + 2(3) = 12 + 6 = 18 \, \text{V} \]

7. **Final answer:**
- The emf of the battery is \( E = 18 \, \text{V} \).