A wire of resistance 0.1 `cm^(-1)` bent to form a square ABCD of side 10 cm. A simailar wire is connected between the corners B and D to form the diagonal BD. Find the effective resistance of this combination between cornaers A and C. If a 2V battery of negligible internal resistance is connected across A and C calculate the total power dissipated.

Options

A. `1Omega,3W`
B. `1Omega, 4W`
C. `2Omega, 3 W`
D. `2 Omega, 4W`

Answer

To solve the problem step by step, we will analyze the circuit formed by the wire and calculate the effective resistance between points A and C, and then find the power dissipated when a 2V battery is connected across A and C.

### Step 1: Calculate the Resistance of Each Side of the Square
The resistance of the wire is given as 0.1 ohm per centimeter, and the length of each side of the square is 10 cm.

\[ R_{\text{side}} = \text{Resistance per cm} \times \text{Length in cm} = 0.1 \, \Omega/\text{cm} \times 10 \, \text{cm} = 1 \, \Omega \]

### Step 2: Identify the Configuration of the Circuit
The square ABCD has a wire connected along the diagonal BD. The resistances along the sides AB, BC, CD, and DA are each 1 ohm. The diagonal BD does not affect the current flow between A and C because it forms a Wheatstone bridge configuration.

### Step 3: Analyze the Wheatstone Bridge Configuration
In a Wheatstone bridge, if the two opposite sides are equal, no current flows through the diagonal. Thus, the effective resistance between points A and C can be calculated by considering only the resistances along the sides AB and CD, and BC and DA.

- Resistance from A to B (R1) = 1 Ω
- Resistance from B to C (R2) = 1 Ω
- Resistance from C to D (R3) = 1 Ω
- Resistance from D to A (R4) = 1 Ω

Since the diagonal BD does not carry any current, we can simplify the circuit to two parallel resistances:

\[ R_{AC} = R_{AB} + R_{BC} \quad \text{and} \quad R_{CD} + R_{DA} \]

### Step 4: Calculate the Effective Resistance
The two paths from A to C are:
1. A to B to C (R1 + R2 = 1 + 1 = 2 Ω)
2. A to D to C (R3 + R4 = 1 + 1 = 2 Ω)

These two paths are in parallel:

\[ \frac{1}{R_{\text{effective}}} = \frac{1}{R_{AB} + R_{BC}} + \frac{1}{R_{CD} + R_{DA}} = \frac{1}{2} + \frac{1}{2} = 1 \]

Thus,

\[ R_{\text{effective}} = 1 \, \Omega \]

### Step 5: Calculate the Power Dissipated
Using the formula for power dissipated in a resistor:

\[ P = \frac{V^2}{R_{\text{effective}}} \]

Substituting the values:

\[ P = \frac{2^2}{1} = \frac{4}{1} = 4 \, \text{W} \]

### Final Answers
- The effective resistance between corners A and C is **1 ohm**.
- The total power dissipated is **4 watts**.