What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected pa

Q: Why do we need to convert the units of resistivity?

The diameter is given in millimeters and the final answer is expected in meters. To maintain consistency and avoid calculation errors, it's best to convert all quantities to standard SI units (meters, ohms) before plugging them into the formulas.

Q: How do we find the cross-sectional area?

The area of a cylindrical wire is given by $A = \pi r^2$. Since the diameter is $2\text{ mm}$, the radius is $1\text{ mm}$ or $10^{-3}\text{ m}$. Squaring this gives $10^{-6}\text{ m}^2$.

Q: What is the formula for parallel resistance?

For two resistors $R_1$ and $R_2$ connected in parallel, the equivalent resistance $R_{eq}$ is given by the reciprocal formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.

Q: Can we solve this without converting to meters first?

Yes, you can keep everything in centimeters and convert the final length from cm to meters at the very end. However, you must be extremely careful to also convert the diameter into centimeters to ensure all units match.

Q: Why is the equivalent resistance of parallel wires less than their individual resistances?

Connecting wires in parallel provides multiple paths for the current to flow. This effectively increases the total cross-sectional area available for conduction, thereby reducing the overall resistance of the system.

Question

What equal length of an iron wire and a copper-nickel alloy wire, each of $2	ext{ mm}$ diameter connected parallel to give an equivalent resistance of $3\Omega$?
(Given, resistivities of iron and copper-nickel alloy wire are $12\mu\Omega	ext{ cm}$ and $51\mu\Omega	ext{ cm}$ respectively)

Options:

(A) 82 m

(B) 97 m

(C) 110 m

(D) 90 m

Sigma Percentile · Accepted Answer

Imagine you are an electrical engineer tasked with designing a specific circuit component. You have two spools of wire: one made of iron and the other of a copper-nickel alloy. Your goal is to cut equal lengths of both wires and connect them in parallel to achieve an exact equivalent resistance of $3\Omega$. This problem is a beautiful exercise in combining the fundamental laws of resistance with parallel circuit theory.

### Visualizing the Circuit

First, let's picture the physical setup. We have two wires, let's call them Wire 1 (Iron) and Wire 2 (Copper-Nickel). They are connected side-by-side, meaning the current splits between them.

Both wires have the same diameter, $d = 2	ext{ mm}$. This means their radius is $r = 1	ext{ mm} = 10^{-3}	ext{ m}$.

The cross-sectional area $A$ for both wires is identical and can be calculated using the formula for the area of a circle:
$$A = \pi r^2 = \pi (10^{-3})^2 = \pi 	imes 10^{-6}	ext{ m}^2$$

### The Importance of Units

Before we jump into the resistance formula, we must address a common trap: mixed units. The resistivities are given in $\mu\Omega	ext{ cm}$. To ensure our final length is in meters, we must convert these resistivities into standard SI units ($\Omega	ext{ m}$).

For the iron wire:
$$ho_1 = 12\mu\Omega	ext{ cm} = 12 	imes 10^{-6}\Omega	ext{ cm} = 12 	imes 10^{-8}\Omega	ext{ m}$$

For the copper-nickel alloy wire:
$$ho_2 = 51\mu\Omega	ext{ cm} = 51 	imes 10^{-6}\Omega	ext{ cm} = 51 	imes 10^{-8}\Omega	ext{ m}$$

### Expressing the Resistances

The resistance of any uniform wire is given by Pouillet's law: $R = \frac{ho l}{A}$. Since we are looking for an equal length $l$ for both wires, we can express their individual resistances as functions of $l$.

For the iron wire:
$$R_1 = \frac{ho_1 l}{A} = \frac{12 	imes 10^{-8} 	imes l}{\pi 	imes 10^{-6}} = \frac{12 	imes 10^{-2} l}{\pi}$$

For the copper-nickel wire:
$$R_2 = \frac{ho_2 l}{A} = \frac{51 	imes 10^{-8} 	imes l}{\pi 	imes 10^{-6}} = \frac{51 	imes 10^{-2} l}{\pi}$$

### The Parallel Combination

When resistors are connected in parallel, their equivalent resistance $R_{eq}$ is found using the reciprocal sum formula:
$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$

We are given that the equivalent resistance must be $3\Omega$. Substituting our expressions for $R_1$ and $R_2$ into the formula, we get:
$$\frac{1}{3} = \frac{\pi}{12 	imes 10^{-2} l} + \frac{\pi}{51 	imes 10^{-2} l}$$

### The Final Calculation

Now, it's just a matter of careful algebra. Let's factor out the common terms to simplify the right side of the equation:
$$\frac{1}{3} = \frac{\pi}{10^{-2} l} \left( \frac{1}{12} + \frac{1}{51} ight)$$

To add the fractions, we find the least common multiple (LCM) of 12 and 51, which is 204.
$$\frac{1}{12} + \frac{1}{51} = \frac{17}{204} + \frac{4}{204} = \frac{21}{204}$$

This fraction simplifies nicely to $\frac{7}{68}$. Substituting this back into our equation:
$$\frac{1}{3} = \frac{100 \pi}{l} \left( \frac{7}{68} ight)$$

Rearranging to solve for $l$:
$$l = 3 	imes 100 \pi 	imes \frac{7}{68} = \frac{2100 \pi}{68}$$

Using the approximation $\pi \approx 3.14$, we find:
$$l \approx \frac{2100 	imes 3.14}{68} \approx 97	ext{ m}$$

Thus, you would need to cut approximately 97 meters of each wire to achieve the desired $3\Omega$ equivalent resistance in parallel. This problem perfectly illustrates how physical dimensions and material properties intertwine to dictate electrical behavior.

Select Answer:

Visualized Solution

$R_{e q}$ of Parallel Wires

Resistivities in SI Units

Resistances $R_{1}$ and $R_{2}$

Parallel Combination

Solving for $l$

Final Calculation

The Way Forward

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Visualizing the Circuit

The Importance of Units

Expressing the Resistances

The Parallel Combination

The Final Calculation

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