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

Answer

To find the e.m.f. of the cell, we can use Ohm's law and the concept of internal resistance. Let's break down the solution step by step.

### Step 1: Write down the known values
- When a resistance \( R_1 = 2 \, \Omega \) is connected, the current \( I_1 = 0.5 \, A \).
- When a resistance \( R_2 = 5 \, \Omega \) is connected, the current \( I_2 = 0.25 \, A \).

### Step 2: Use Ohm's law
According to Ohm's law, the relationship between e.m.f. (E), current (I), resistance (R), and internal resistance (r) can be expressed as:
\[ I = \frac{E}{R + r} \]

### Step 3: Set up equations for both cases
1. For the first case:
\[ I_1 = \frac{E}{R_1 + r} \implies 0.5 = \frac{E}{2 + r} \quad \text{(Equation 1)} \]

2. For the second case:
\[ I_2 = \frac{E}{R_2 + r} \implies 0.25 = \frac{E}{5 + r} \quad \text{(Equation 2)} \]

### Step 4: Rearranging the equations
From Equation 1:
\[ E = 0.5(2 + r) \implies E = 1 + 0.5r \quad \text{(Equation 3)} \]

From Equation 2:
\[ E = 0.25(5 + r) \implies E = 1.25 + 0.25r \quad \text{(Equation 4)} \]

### Step 5: Set Equations 3 and 4 equal to each other
Since both expressions equal E, we can set them equal:
\[ 1 + 0.5r = 1.25 + 0.25r \]

### Step 6: Solve for r
Rearranging gives:
\[ 0.5r - 0.25r = 1.25 - 1 \]
\[ 0.25r = 0.25 \implies r = 1 \, \Omega \]

### Step 7: Substitute r back into one of the equations to find E
Using Equation 3:
\[ E = 1 + 0.5(1) = 1 + 0.5 = 1.5 \, V \]

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

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