A uniform wire of resistance `9 Omega` is cut into 3 equal parts. They are connected in form of equilateral triangle `ABC`. A cell of e.m.f. `2V` and negligible internal resistance is connected across `B` and `C`. Potential difference across `AB` is

Options

A. 1V
B. 2V
C. 3V
D. 0.5V

Answer

To find the potential difference across AB in the given circuit, we will follow these steps:

### Step 1: Determine the Resistance of Each Segment
The total resistance of the uniform wire is given as \(9 \, \Omega\). When this wire is cut into 3 equal parts, the resistance of each part is:
\[ R_{\text{each}} = \frac{9 \, \Omega}{3} = 3 \, \Omega \]

### Step 2: Analyze the Circuit Configuration
The three resistances are connected to form an equilateral triangle (ABC). The resistances are:
- \(R_{AB} = 3 \, \Omega\)
- \(R_{BC} = 3 \, \Omega\)
- \(R_{CA} = 3 \, \Omega\)

### Step 3: Identify the Connection of the Battery
A cell with an EMF of \(2 \, V\) is connected across points B and C. This means that the potential difference is applied across the resistance \(R_{BC}\).

### Step 4: Calculate the Equivalent Resistance
In the triangle configuration, the resistances \(R_{AB}\) and \(R_{CA}\) are in parallel with respect to the resistance \(R_{BC}\).

The equivalent resistance \(R_{eq}\) can be calculated as follows:
1. The two resistances \(R_{AB}\) and \(R_{CA}\) are in series with each other, and their equivalent resistance \(R_{AC}\) is:
\[ R_{AC} = R_{AB} + R_{CA} = 3 \, \Omega + 3 \, \Omega = 6 \, \Omega \]
2. Now, this equivalent resistance \(R_{AC}\) is in parallel with \(R_{BC}\):
\[ \frac{1}{R_{eq}} = \frac{1}{R_{BC}} + \frac{1}{R_{AC}} = \frac{1}{3} + \frac{1}{6} \]
\[ \frac{1}{R_{eq}} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]
\[ R_{eq} = 2 \, \Omega \]

### Step 5: Calculate the Total Current from the Battery
Using Ohm's law, the total current \(I\) flowing from the battery can be calculated as:
\[ I = \frac{V}{R_{eq}} = \frac{2 \, V}{2 \, \Omega} = 1 \, A \]

### Step 6: Use Current Division to Find Current in AB
Let \(I_1\) be the current through \(R_{AB}\) and \(I_2\) be the current through \(R_{CA}\). Using the current division rule:
\[ I_1 = I \cdot \frac{R_{CA}}{R_{AB} + R_{CA}} = 1 \cdot \frac{3}{3 + 3} = 1 \cdot \frac{3}{6} = \frac{1}{2} \, A \]

### Step 7: Calculate the Potential Difference Across AB
The potential difference \(V_{AB}\) across \(R_{AB}\) can be calculated using Ohm's law:
\[ V_{AB} = I_1 \cdot R_{AB} = \frac{1}{2} \, A \cdot 3 \, \Omega = \frac{3}{2} \, V = 1.5 \, V \]

### Final Answer
The potential difference across \(AB\) is \(1.5 \, V\).

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