A certain length of a uniform wire of resistance `12Omega` is bent into a circle and two points, a quarter of circumference apart, are connected to a battery of emf 4 V and internal resistance `1Omega`. Find the current in the different parts of the circuit.

Answer

To solve the problem, we need to find the current in different parts of the circuit formed by a uniform wire bent into a circle, connected to a battery. Here’s a step-by-step solution:

### Step 1: Understand the Circuit Configuration
The wire has a total resistance of \( R = 12 \, \Omega \) and is bent into a circle. When two points on the circle are connected to a battery, which is \( \frac{1}{4} \) of the circumference apart, we can visualize the circuit.

### Step 2: Determine the Lengths of the Wire Segments
The circumference of the circle is given by \( C = 2\pi r \). Since the wire is uniform, the resistance is proportional to the length. The segment connected to the battery is \( \frac{1}{4} \) of the circumference, and the remaining segment is \( \frac{3}{4} \).

- Length of segment 1 (connected to the battery): \( L_1 = \frac{L}{4} \)
- Length of segment 2: \( L_2 = \frac{3L}{4} \)

### Step 3: Calculate the Resistances of the Segments
Using the relationship \( R = \rho \frac{L}{A} \), we can find the resistances of the segments:

- Resistance of segment 1 (\( R_1 \)):
\[ R_1 = \frac{12 \, \Omega}{4} = 3 \, \Omega \]

- Resistance of segment 2 (\( R_2 \)):
\[ R_2 = \frac{3}{4} \times 12 \, \Omega = 9 \, \Omega \]

### Step 4: Analyze the Circuit with the Battery
The battery has an emf of \( 4 \, V \) and an internal resistance of \( 1 \, \Omega \). The total resistance in the circuit is the sum of the resistances of the segments and the internal resistance of the battery.

### Step 5: Calculate the Equivalent Resistance
The equivalent resistance \( R_{AB} \) between points A and B (where the battery is connected) is given by:
\[ R_{AB} = R_1 + R_2 = 3 \, \Omega + 9 \, \Omega = 12 \, \Omega \]
Adding the internal resistance of the battery:
\[ R_{eq} = R_{AB} + R_{internal} = 12 \, \Omega + 1 \, \Omega = 13 \, \Omega \]

### Step 6: Calculate the Total Current from the Battery
Using Ohm's law, the total current \( I \) from the battery is:
\[ I = \frac{V}{R_{eq}} = \frac{4 \, V}{13 \, \Omega} = \frac{4}{13} \, A \]

### Step 7: Determine the Current Distribution
Since the resistances are in series, the current divides in the ratio of the resistances. The current \( I_1 \) through \( R_1 \) and \( I_2 \) through \( R_2 \) can be calculated using the current division rule.

Using the ratio of the resistances:
\[ \frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{9}{3} = 3 \]
Let \( I_1 = 3x \) and \( I_2 = x \). Then:
\[ I_1 + I_2 = I \implies 3x + x = \frac{4}{13} \implies 4x = \frac{4}{13} \implies x = \frac{1}{13} \]
Thus:
\[ I_1 = 3x = \frac{3}{13} \, A, \quad I_2 = x = \frac{1}{13} \, A \]

### Final Answer
- Current through segment 1 (\( I_1 \)): \( \frac{3}{13} \, A \)
- Current through segment 2 (\( I_2 \)): \( \frac{1}{13} \, A \)