To solve the problem of finding the effective resistance between points A and C in the given wire configuration, we need to analyze the arrangement of the wires and apply some principles of electrical circuits. Let's break it down step by step.
Understanding the Wire Configuration
We have a square ABCD with each side measuring 1 cm. The resistance of the wire is given as 0.1 ohm per cm. Since the wire is bent to form a square, we can calculate the resistance of each side of the square:
- Resistance of one side (AB, BC, CD, or DA) = 0.1 ohm/cm × 1 cm = 0.1 ohm
Calculating the Resistance of the Diagonal
Next, we need to consider the wire connecting points B and D. This wire is also made of the same material and has the same resistance per unit length. The length of the diagonal BD can be calculated using the Pythagorean theorem:
- Length of diagonal BD = √(1² + 1²) = √2 cm
- Resistance of diagonal BD = 0.1 ohm/cm × √2 cm = 0.1√2 ohm
Effective Resistance Between Points A and C
Now, we can analyze the circuit formed by the square and the diagonal. The resistances can be represented as follows:
- AB = 0.1 ohm
- BC = 0.1 ohm
- CD = 0.1 ohm
- DA = 0.1 ohm
- BD = 0.1√2 ohm
When we look at the path from A to C, we see that there are two parallel paths:
- Path 1: A → B → C (which includes AB and BC)
- Path 2: A → D → C (which includes AD and CD) and the diagonal BD
For Path 1 (A to C via B):
- R1 = R_AB + R_BC = 0.1 + 0.1 = 0.2 ohm
For Path 2 (A to C via D):
- R2 = R_AD + R_BD + R_CD = 0.1 + 0.1√2 + 0.1 = 0.2 + 0.1√2 ohm
Finding the Total Resistance
Now, since R1 and R2 are in parallel, we can find the equivalent resistance (R_eq) using the formula for resistances in parallel:
1/R_eq = 1/R1 + 1/R2
Substituting the values:
1/R_eq = 1/0.2 + 1/(0.2 + 0.1√2)
Calculating R_eq gives us:
R_eq ≈ 0.1 ohm
Power Dissipation Calculation
Now that we have the effective resistance, we can calculate the power dissipated when a 2 V battery is connected across points A and C. The power (P) can be calculated using the formula:
P = V² / R_eq
Substituting the values:
P = (2 V)² / 0.1 ohm = 4 W
Final Thoughts
The effective resistance between points A and C is approximately 0.1 ohm, and when a 2 V battery is connected, the power dissipated in the circuit is 4 watts. This example illustrates how to analyze a circuit with both series and parallel components effectively.
