When a resistance of 2 ohm is connected across terminals of a cell, the current is `0.5 A`. When the resistance is increased to 5 ohm, the current is `0.25 A` The e.m.f. of the cell is

Options

A. `1.0 V`
B. `1.5 V`
C. `2.0 V`
D. `2.5 V`

Explanation

To find the e.m.f. of the cell, we will use the information given in the problem and apply Ohm's law.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
We have a cell with an internal resistance \( r \) and an external resistance \( R \). The e.m.f. of the cell is denoted as \( E \).

2. **Using Ohm's Law**:
According to Ohm's law, the current \( I \) flowing through the circuit can be expressed as:
\[ I = \frac{E}{R + r} \]

3. **Setting Up the First Condition**:
When \( R = 2 \, \Omega \), the current \( I = 0.5 \, A \):
\[ 0.5 = \frac{E}{2 + r} \quad \text{(Equation 1)} \]

4. **Setting Up the Second Condition**:
When \( R = 5 \, \Omega \), the current \( I = 0.25 \, A \):
\[ 0.25 = \frac{E}{5 + r} \quad \text{(Equation 2)} \]

5. **Manipulating Equation 2**:
Multiply Equation 2 by 2 to facilitate comparison:
\[ 0.5 = \frac{2E}{5 + r} \quad \text{(Equation 3)} \]

6. **Equating the Two Expressions for Current**:
From Equations 1 and 3, we can set them equal to each other:
\[ \frac{E}{2 + r} = \frac{2E}{5 + r} \]

7. **Cross-Multiplying**:
Cross-multiplying gives:
\[ E(5 + r) = 2E(2 + r) \]

8. **Expanding Both Sides**:
Expanding both sides leads to:
\[ 5E + Er = 4E + 2Er \]

9. **Rearranging the Equation**:
Rearranging gives:
\[ 5E - 4E = 2Er - Er \]
\[ E = Er \]

10. **Solving for Internal Resistance**:
Rearranging gives:
\[ E = r \]
Substitute \( r \) back into either Equation 1 or Equation 2 to find \( r \).

11. **Substituting \( r \) into Equation 1**:
Using Equation 1:
\[ 0.5 = \frac{E}{2 + r} \]
Substitute \( r = 1 \):
\[ 0.5 = \frac{E}{2 + 1} \]
\[ 0.5 = \frac{E}{3} \]
\[ E = 0.5 \times 3 = 1.5 \, V \]

### Final Answer:
The e.m.f. of the cell is \( E = 1.5 \, V \).

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