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 find the EMF of the battery, we will analyze the two cases provided in the question step by step.

### Step 1: Analyze Case 1
In the first case, a current of 2 A flows from the negative to the positive terminal, resulting in a potential difference of 12 V across the battery.

- Let \( e \) be the EMF of the battery.
- Let \( r \) be the internal resistance of the battery.

Using Ohm's law, the potential difference across the internal resistance can be expressed as:
\[ V = I \cdot r \]
For a current of 2 A:
\[ V = 2r \]

Applying Kirchhoff's loop rule, we can write:
\[ e - 2r - 12 = 0 \]
Rearranging gives us:
\[ e = 12 + 2r \quad \text{(Equation 1)} \]

### Step 2: Analyze Case 2
In the second case, a current of 3 A flows in the opposite direction, resulting in a potential difference of 15 V across the battery.

Using Ohm's law again, for a current of 3 A:
\[ V = 3r \]

Applying Kirchhoff's loop rule for this case gives us:
\[ e - 3r + 15 = 0 \]
Rearranging gives us:
\[ e = 15 + 3r \quad \text{(Equation 2)} \]

### Step 3: Equate the Two Equations
Now we have two equations for \( e \):
1. \( e = 12 + 2r \)
2. \( e = 15 + 3r \)

Setting these two equations equal to each other:
\[ 12 + 2r = 15 + 3r \]

### Step 4: Solve for \( r \)
Rearranging the equation:
\[ 12 - 15 = 3r - 2r \]
\[ -3 = r \]
Thus, we find:
\[ r = 3 \, \text{ohms} \]

### Step 5: Substitute \( r \) Back to Find \( e \)
Now, substitute \( r \) back into either Equation 1 or Equation 2 to find \( e \). Using Equation 1:
\[ e = 12 + 2(3) \]
\[ e = 12 + 6 = 18 \, \text{V} \]

### Final Answer
The EMF of the battery is \( \text{18 V} \).

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