A current of `2 A` flows through a `2 Omega` resistor when connected across a battery. The same battery supplies a current of `0.5 A` when connected across a `9 Omega` resistor. The internal resistance of the battery is

Options

A. ` 1 Omega`
B. `0.5Omega`
C. `1//3 Omega`
D. `-1//4 Omega`

Answer

To find the internal resistance of the battery, 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 equations for both scenarios.

1. **For the first scenario** (2 Ω resistor with a current of 2 A):
\[ I_1 = \frac{E}{R_1 + r} \]
where \( I_1 = 2 \, \text{A} \), \( R_1 = 2 \, \Omega \), and \( r \) is the internal resistance.

Rearranging gives:
\[ E = I_1(R_1 + r) = 2(2 + r) = 4 + 2r \]
This is our **Equation 1**.

2. **For the second scenario** (9 Ω resistor with a current of 0.5 A):
\[ I_2 = \frac{E}{R_2 + r} \]
where \( I_2 = 0.5 \, \text{A} \) and \( R_2 = 9 \, \Omega \).

Rearranging gives:
\[ E = I_2(R_2 + r) = 0.5(9 + r) = 4.5 + 0.5r \]
This is our **Equation 2**.

### Step 2: Set the two equations equal to each other.

Since both equations represent the same battery EMF \( E \), we can set them equal:
\[ 4 + 2r = 4.5 + 0.5r \]

### Step 3: Solve for the internal resistance \( r \).

1. Rearranging the equation:
\[ 2r - 0.5r = 4.5 - 4 \]
\[ 1.5r = 0.5 \]

2. Dividing both sides by 1.5:
\[ r = \frac{0.5}{1.5} = \frac{1}{3} \, \Omega \]

### Final Answer:

The internal resistance of the battery is \( \frac{1}{3} \, \Omega \).

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