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

Explanation

To find the internal resistance of the battery, we can analyze the given information step by step.

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

1. When a `2 Ω` resistor is connected to the battery, a current of `2 A` flows through it.
- Using Ohm's law, we can express this as:
\[ I = \frac{E}{R + r} \]
where \( E \) is the electromotive force (emf) of the battery, \( R \) is the external resistance, and \( r \) is the internal resistance of the battery.
- For this case:
\[ 2 = \frac{E}{2 + r} \quad \text{(Equation 1)} \]

2. When a `9 Ω` resistor is connected to the battery, a current of `0.5 A` flows through it.
- Similarly, we can express this as:
\[ 0.5 = \frac{E}{9 + r} \quad \text{(Equation 2)} \]

### Step 2: Rearranging the equations

From Equation 1:
\[ E = 2(2 + r) = 4 + 2r \quad \text{(1)} \]

From Equation 2:
\[ E = 0.5(9 + r) = 4.5 + 0.5r \quad \text{(2)} \]

### Step 3: Set the two expressions for E equal to each other

Setting the two expressions for \( E \) from Equations (1) and (2) equal:
\[ 4 + 2r = 4.5 + 0.5r \]

### Step 4: Solve for r

Now, we can solve for \( r \):
\[ 4 + 2r - 0.5r = 4.5 \]
\[ 4 + 1.5r = 4.5 \]
\[ 1.5r = 4.5 - 4 \]
\[ 1.5r = 0.5 \]
\[ r = \frac{0.5}{1.5} = \frac{1}{3} \, \Omega \]

### Conclusion

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

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