To solve this problem, we need to analyze the arrangement of the wires and calculate the effective resistance between points A and C. Let's break it down step by step.
Understanding the Wire Configuration
We have a square wire ABCD with each side measuring 10 cm. The resistance of the wire is given as 0.1 ohm per cm. First, we need to calculate the total resistance of the wire forming the square.
Calculating the Resistance of the Square
The total length of the wire forming the square is the perimeter, which can be calculated as:
- Perimeter = 4 × side = 4 × 10 cm = 40 cm
Now, we can find the total resistance of the square wire:
- Resistance of the square = Resistance per cm × Length = 0.1 ohm/cm × 40 cm = 4 ohms
Resistance of the Diagonal Wire
Next, we need to consider the wire connected between corners B and D, which forms the diagonal BD. The length of the diagonal can be calculated using the Pythagorean theorem:
- Length of diagonal BD = √(side² + side²) = √(10² + 10²) = √200 = 10√2 cm
Now, we can calculate the resistance of this diagonal wire:
- Resistance of diagonal BD = 0.1 ohm/cm × 10√2 cm = 1√2 ohms ≈ 1.414 ohms
Combining the Resistances
Now, we need to determine how these resistances are connected. The square wire ABCD has points A and C connected to the diagonal BD. The effective resistance between A and C can be found by recognizing that the two paths (through the square and through the diagonal) are in parallel.
Calculating the Effective Resistance
The resistance of the path through the square wire (AC) is 4 ohms, and the resistance of the diagonal (BD) is approximately 1.414 ohms. The formula for the effective resistance \( R_{eff} \) of two resistors in parallel \( R_1 \) and \( R_2 \) is given by:
- \( \frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2} \)
Substituting the values:
- \( \frac{1}{R_{eff}} = \frac{1}{4} + \frac{1}{1.414} \)
- \( \frac{1}{R_{eff}} = 0.25 + 0.707 \approx 0.957 \)
- \( R_{eff} \approx \frac{1}{0.957} \approx 1.05 \text{ ohms} \)
Finding the Power Across A and C
Now that we have the effective resistance, we can calculate the power when a 2 V battery is connected across points A and C. The power \( P \) can be calculated using the formula:
- \( P = \frac{V^2}{R_{eff}} \)
Substituting the values:
- \( P = \frac{2^2}{1.05} \approx \frac{4}{1.05} \approx 3.81 \text{ watts} \)
Summary of Results
To summarize:
- The effective resistance between points A and C is approximately 1.05 ohms.
- The power dissipated across A and C when a 2 V battery is connected is approximately 3.81 watts.
This analysis shows how to approach problems involving resistances in complex configurations, using both series and parallel combinations effectively. If you have any further questions or need clarification on any step, feel free to ask!
