Two wires of same length and thickness having specific resistances 6 ohm-cm and 3 ohm-cm respectivel

Q: Why does the area become $2A$ for the equivalent wire?

When two wires of the same thickness are placed side-by-side in a parallel circuit, the total cross-sectional area available for the current to flow is the sum of their individual areas. It's like opening a second identical lane on a highway.

Q: Why doesn't the length change in a parallel connection?

In a parallel connection, the current travels the exact same distance from the starting junction to the ending junction through either branch. Therefore, the effective length of the composite system remains $l$.

Q: Can I just take the average of the two resistivities?

No, resistivity doesn't average linearly in a parallel configuration. The relationship is harmonic, similar to how parallel resistors behave, but it is also scaled by the area factor, leading to the $\frac{\rho}{2}$ term.

Q: What if the wires had different lengths?

If the lengths were different, the concept of a single 'effective resistivity' for a uniform equivalent wire would be much more complex to define, as the resulting geometry wouldn't behave like a simple uniform cylinder.

Q: Is effective resistivity a real physical property of a material?

No, it's a mathematical construct. It represents the combination of two different materials as if they were replaced by a single, uniform material that provides the exact same electrical resistance.

Question

Two wires of same length and thickness having specific resistances $6\Omega$- cm and $3\Omega$-cm respectively are connected in parallel. The effective resistivity is $\rho$ $\Omega$-cm. The value of $\rho$ to the nearest integer, is ............... .

Sigma Percentile · Accepted Answer

### The Physical Setup
Imagine two identical wires, having the exact same length $l$ and cross-sectional area $A$, but made of entirely different materials. Because they are made of different materials, they possess different specific resistances (or resistivities), denoted as $ho_1$ and $ho_2$.

When we connect these two wires in parallel, we are essentially creating a new, composite wire. To find the effective resistivity $ho$ of this composite wire, we must first understand how its physical dimensions behave.

### The Dimensional Catch
In a parallel connection, the current splits and travels through both wires simultaneously. Since the wires are placed side-by-side, the total distance the current must travel remains unchanged. Therefore, the equivalent length is simply $L_{eq} = l$.

However, the cross-sectional area available for the current to flow has now doubled! The equivalent wire has an area $A_{eq} = A + A = 2A$. This is a crucial conceptual step. If you miss this, the entire derivation falls apart.

### The Master Equation
We know the resistance of any uniform wire is given by $R = \frac{ho l}{A}$.

For our individual wires, we have:
$R_1 = \frac{ho_1 l}{A}$
$R_2 = \frac{ho_2 l}{A}$

For our equivalent composite wire, the resistance is:
$R_{eq} = \frac{ho l}{2A}$

We also know the fundamental law for parallel resistors:
$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

### The Derivation
Let's substitute our physical resistance formulas into the parallel combination equation:

$$\frac{ho l}{2A} = \frac{\left(\frac{ho_1 l}{A}ight) \left(\frac{ho_2 l}{A}ight)}{\frac{ho_1 l}{A} + \frac{ho_2 l}{A}}$$

I know this looks like a terrifying fraction, but let's take a breath. Notice that every single term contains the factor $\frac{l}{A}$. We can factor it out from the denominator and cancel it beautifully with the numerator and the left-hand side.

After the dust settles, we are left with a remarkably elegant relation:

$$\frac{ho}{2} = \frac{ho_1 ho_2}{ho_1 + ho_2}$$

### Final Calculation
We are given the specific resistances $ho_1 = 6 \, \Omega	ext{-cm}$ and $ho_2 = 3 \, \Omega	ext{-cm}$. Let's plug these numbers into our derived relation:

$$\frac{ho}{2} = \frac{6 	imes 3}{6 + 3}$$

$$\frac{ho}{2} = \frac{18}{9}$$

$$\frac{ho}{2} = 2$$

Multiplying both sides by 2, we arrive at our final answer:

**$ho = 4 \, \Omega	ext{-cm}$**

Animated Solution for Physics - Current Electricity: Two wires of same length and thickness having specific resistances $6Ω$ - cm and $3Ω$ -cm respectively are connected in parallel. The effective resistivity is $ρ$ $Ω$ -cm. The value of $ρ$ to the nearest integer, is ............... .

Enter Numerical Value:

Visualized Solution

$R = \frac{ρl}{A}$

$R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$

$A_{e q} = 2 A$

$R_{e q} = \frac{ρl}{2 A}$

$Substitution$

$Simplification$

$Calculation$

$ρ = 4$

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

The Physical Setup Imagine two identical wires, having the exact same length $l$ and cross-sectional area $A$ , but made of entirely different materials

The Dimensional Catch In a parallel connection, the current splits and travels through both wires simultaneously

The Master Equation

The Derivation

Final Calculation We are given the specific resistances $ρ_{1} = 6 Ω -cm$ and $ρ_{2} = 3 Ω -cm$

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=Aρl​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​=R1​+R2​R1​R2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Aeq​=2A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​=2Aρl​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρ=4