Question

The length of a wire of $10 \, \Omega$ resistance is three times the length on stretching it. Now the wire is cut into three equal parts and then they are joined in an electrical circuit as shown in the figure. Find out the total resistance of the combination between A and B.

Hint:

Stretching a wire increases its resistance proportional to the square of stretching factor.

Answer 1

Step 1: Effect of stretching on resistance.
Resistance of a wire: \[ R = \rho \frac{L}{A}. \] If the wire is stretched to 3 times its original length, then $L' = 3L$ and volume remains constant. \[ A' = \frac{A}{3}. \] So, \[ R' = \rho \frac{3L}{A/3} = 9 \cdot \rho \frac{L}{A} = 9R. \]
Step 2: New resistance of wire.
Given original resistance = $10 \, \Omega$. \[ R' = 9 \times 10 = 90 \, \Omega. \]
Step 3: Cutting into 3 equal parts.
When cut into 3 equal parts: \[ R_{part} = \frac{90}{3} = 30 \, \Omega \, \text{each}. \]
Step 4: Analyze circuit.
- The top two resistors (each $30 \, \Omega$) are in series: \[ R_{top} = 30 + 30 = 60 \, \Omega. \] - The bottom resistor is $30 \, \Omega$. - These two branches are in parallel.
Step 5: Equivalent resistance.
\[ \frac{1}{R_{AB}} = \frac{1}{R_{top}} + \frac{1}{R_{bottom}} = \frac{1}{60} + \frac{1}{30}. \] \[ \frac{1}{R_{AB}} = \frac{1 + 2}{60} = \frac{3}{60} = \frac{1}{20}. \] \[ R_{AB} = 20 \, \Omega. \] ⚠ Correction: Let’s carefully recalc — Wait: $R_{top} = 60 \, \Omega$, $R_{bottom} = 30 \, \Omega$. So, \[ R_{eq} = \frac{(60)(30)}{60 + 30} = \frac{1800}{90} = 20 \, \Omega. \] Final answer: $R_{AB} = 20 \, \Omega$.
Step 6: Conclusion.
The total resistance between A and B is $20 \, \Omega$.