When a resistance of 2 ohm is placed across a battery the current is 1 A and when the resistance across the terminals is 17 ohm, the current is 0.25A. The emf of the battery is

Options

A. 4.5 V
B. 5V
C. 3V
D. 6V

Explanation

To find the emf of the battery, we can use Ohm's law and the concept of internal resistance. Let's denote the emf of the battery as \( E \) and the internal resistance as \( r \).

### Step 1: Set up the equations based on the given information.

1. When a resistance of \( 2 \, \Omega \) is connected, the current is \( 1 \, A \).
\[ E = I_1 (R + r) \quad \text{(where \( R = 2 \, \Omega \) and \( I_1 = 1 \, A \))} \]
\[ E = 1(2 + r) \quad \Rightarrow \quad E = 2 + r \quad \text{(Equation 1)} \]

2. When a resistance of \( 17 \, \Omega \) is connected, the current is \( 0.25 \, A \).
\[ E = I_2 (R + r) \quad \text{(where \( R = 17 \, \Omega \) and \( I_2 = 0.25 \, A \))} \]
\[ E = 0.25(17 + r) \quad \Rightarrow \quad E = 4.25 + 0.25r \quad \text{(Equation 2)} \]

### Step 2: Equate the two expressions for \( E \).

From Equation 1:
\[ E = 2 + r \]

From Equation 2:
\[ E = 4.25 + 0.25r \]

Setting them equal to each other:
\[ 2 + r = 4.25 + 0.25r \]

### Step 3: Solve for \( r \).

Rearranging the equation:
\[ r - 0.25r = 4.25 - 2 \]
\[ 0.75r = 2.25 \]
\[ r = \frac{2.25}{0.75} = 3 \, \Omega \]

### Step 4: Substitute \( r \) back to find \( E \).

Using \( r = 3 \, \Omega \) in Equation 1:
\[ E = 2 + r = 2 + 3 = 5 \, V \]

### Final Answer:
The emf of the battery is \( 5 \, V \).

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