the current in a circuit containing a battery connected to `2Omega` resistance is `0.9 A`. When ana additional resistance of `8Omega` is connected to the same battery the current observed in the circuit is `0.3A`. Then the internal resistance of the battery is `xOmega`. Find value of x.

Answer

To find the internal resistance \( x \) of the battery, we can use Ohm's law and the information provided in the problem. Let's break it down step by step.

### Step 1: Analyze the first case
In the first scenario, the battery is connected to a \( 2 \Omega \) resistor, and the current flowing through the circuit is \( 0.9 A \).

Using Ohm's law, we can express the voltage of the battery \( E \) as:
\[ E = I \cdot (R + r) \]
where \( I \) is the current, \( R \) is the external resistance, and \( r \) is the internal resistance of the battery.

For the first case:
- \( I = 0.9 A \)
- \( R = 2 \Omega \)

So, we can write:
\[ E = 0.9 \cdot (2 + x) \]
This simplifies to:
\[ E = 1.8 + 0.9x \quad \text{(Equation 1)} \]

### Step 2: Analyze the second case
In the second scenario, an additional \( 8 \Omega \) resistor is connected, making the total external resistance \( 10 \Omega \) (since \( 2 + 8 = 10 \)), and the current is now \( 0.3 A \).

Using the same formula:
\[ E = I \cdot (R + r) \]
For the second case:
- \( I = 0.3 A \)
- \( R = 10 \Omega \)

So, we can write:
\[ E = 0.3 \cdot (10 + x) \]
This simplifies to:
\[ E = 3 + 0.3x \quad \text{(Equation 2)} \]

### Step 3: Set the equations equal to each other
Since both equations represent the same battery voltage \( E \), we can set them equal to each other:
\[ 1.8 + 0.9x = 3 + 0.3x \]

### Step 4: Solve for \( x \)
Now, we will rearrange the equation to isolate \( x \):
\[ 1.8 + 0.9x - 0.3x = 3 \]
\[ 1.8 + 0.6x = 3 \]
Subtract \( 1.8 \) from both sides:
\[ 0.6x = 3 - 1.8 \]
\[ 0.6x = 1.2 \]
Now, divide both sides by \( 0.6 \):
\[ x = \frac{1.2}{0.6} = 2 \]

### Conclusion
The internal resistance of the battery is \( x = 2 \Omega \).

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