To solve the problem step by step, we will follow these steps:
### Step 1: Understand the Configuration
We have a uniform wire of resistance \( R = 12 \, \Omega \) bent into a circle. When bent into a circle, the total resistance remains the same, but we can analyze the circuit based on the segments created.
### Step 2: Determine the Resistance of Each Segment
Since the wire is bent into a circle, the circumference is divided into four equal parts. The resistance of each quarter circumference (arc) can be calculated as follows:
\[
R_{\text{quarter}} = \frac{R}{4} = \frac{12 \, \Omega}{4} = 3 \, \Omega
\]
Thus, the resistance between points A and B (which are a quarter circumference apart) is \( 3 \, \Omega \).
### Step 3: Calculate the Resistance of the Remaining Wire
The remaining part of the wire, which is three-quarters of the circumference, will have a resistance:
\[
R_{\text{remaining}} = R - R_{\text{quarter}} = 12 \, \Omega - 3 \, \Omega = 9 \, \Omega
\]
### Step 4: Analyze the Circuit
The circuit consists of:
- A \( 3 \, \Omega \) resistance (between points A and B)
- A \( 9 \, \Omega \) resistance (the remaining part of the wire)
- A battery with an EMF of \( 4 \, V \) and an internal resistance of \( 1 \, \Omega \)
### Step 5: Find the Equivalent Resistance
The \( 3 \, \Omega \) and \( 9 \, \Omega \) resistances are in parallel. The equivalent resistance \( R_{\text{eq}} \) of two resistors in parallel is given by:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3} + \frac{1}{9}
\]
Calculating this gives:
\[
\frac{1}{R_{\text{eq}}} = \frac{3 + 1}{9} = \frac{4}{9} \implies R_{\text{eq}} = \frac{9}{4} \, \Omega
\]
### Step 6: Add Internal Resistance
Now, we add the internal resistance of the battery:
\[
R_{\text{total}} = R_{\text{eq}} + R_{\text{internal}} = \frac{9}{4} + 1 = \frac{9}{4} + \frac{4}{4} = \frac{13}{4} \, \Omega
\]
### Step 7: Calculate the Total Current
Using Ohm's law \( V = IR \), we can find the total current \( I \) flowing through the circuit:
\[
I = \frac{V}{R_{\text{total}}} = \frac{4 \, V}{\frac{13}{4} \, \Omega} = \frac{4 \times 4}{13} = \frac{16}{13} \, A
\]
### Step 8: Calculate the Current in Each Segment
Using the current division rule, we can find the current through each resistor:
1. For the \( 3 \, \Omega \) resistor (let's call the current through it \( I_1 \)):
\[
I_1 = I \cdot \frac{R_{\text{remaining}}}{R_{\text{quarter}} + R_{\text{remaining}}} = \frac{16}{13} \cdot \frac{9}{3 + 9} = \frac{16}{13} \cdot \frac{9}{12} = \frac{12}{13} \, A
\]
2. For the \( 9 \, \Omega \) resistor (let's call the current through it \( I_2 \)):
\[
I_2 = I - I_1 = \frac{16}{13} - \frac{12}{13} = \frac{4}{13} \, A
\]
### Final Answers
- Current through the \( 3 \, \Omega \) resistor \( I_1 = \frac{12}{13} \, A \)
- Current through the \( 9 \, \Omega \) resistor \( I_2 = \frac{4}{13} \, A \)
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