Two metallic wires made from copper have same length but the radius of wire 1 is half of that of wire 2. The resistance of wire 1 is R. If both the wires are joined together in series, the total resistance becomes

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Prepp Team · Accepted Answer

Understanding Electrical Resistance in Wires

Electrical resistance is a measure of how much a material opposes the flow of electric current. For a uniform conductor, the resistance ($R$) depends on its material, length ($L$), and cross-sectional area ($A$). The relationship is given by the formula:

$R = \rho \frac{L}{A}$

Here, $\rho$ (rho) is the resistivity of the material, which is a property specific to the material (like copper, aluminum, etc.) at a given temperature.

For a wire with a circular cross-section of radius $r$, the area $A = \pi r^2$. So the formula becomes:

$R = \rho \frac{L}{\pi r^2}$

Analyzing the Two Copper Wires

We are given two metallic wires made from copper. This means the resistivity ($\rho$) is the same for both wires. They also have the same length ($L$). The only difference is their radius.

Wire 1: Radius is $r_1$, Resistance is $R_1 = R$.
Wire 2: Radius is $r_2$. We are told the radius of wire 1 is half of that of wire 2. So, $r_1 = \frac{1}{2} r_2$, which means $r_2 = 2r_1$.

Calculating the Resistance of Each Wire

Using the formula $R = \rho \frac{L}{\pi r^2}$:

For Wire 1:

$R_1 = \rho \frac{L}{\pi r_1^2}$

We are given that $R_1 = R$. So, $R = \rho \frac{L}{\pi r_1^2}$.

For Wire 2:

$R_2 = \rho \frac{L}{\pi r_2^2}$

Substitute $r_2 = 2r_1$ into the equation for $R_2$:

$R_2 = \rho \frac{L}{\pi (2r_1)^2} = \rho \frac{L}{\pi (4r_1^2)}$

We can rewrite this as:

$R_2 = \frac{1}{4} \left(\rho \frac{L}{\pi r_1^2}\right)$

Notice that the term in the parenthesis is the expression for $R_1$ (which is equal to $R$).

Therefore, $R_2 = \frac{1}{4} R$.

So, the resistance of wire 2 is one-fourth the resistance of wire 1.

Total Resistance in Series Combination

When two or more resistors are joined in series, the total resistance ($R_{total}$) is the sum of their individual resistances.

$R_{total} = R_1 + R_2$

We know $R_1 = R$ and we calculated $R_2 = \frac{1}{4} R$.

Substituting these values:

$R_{total} = R + \frac{1}{4} R$

To add these terms, find a common denominator:

$R_{total} = \frac{4}{4} R + \frac{1}{4} R = \left(\frac{4+1}{4}\right) R$

$R_{total} = \frac{5}{4} R$

Thus, when the two wires are joined in series, the total resistance is $\frac{5}{4}R$.

Revision Table: Key Concepts

Concept	Description	Formula
Electrical Resistance ($R$)	Opposition to current flow	$R = \rho \frac{L}{A}$
Resistivity ($\rho$)	Material property indicating resistance	Units: Ohm-meter ($\Omega \cdot m$)
Cross-sectional Area ($A$)	Area of the conductor's face perpendicular to current flow	For wire: $A = \pi r^2$
Series Combination	Components connected end-to-end	$R_{total} = R_1 + R_2 + ...$

Additional Information: Factors Affecting Resistance and Conductivity

The resistance of a wire depends on several factors:

Material: Different materials have different resistivities. Conductors (like copper, silver) have low resistivity, while insulators (like rubber, glass) have high resistivity. Resistivity also typically changes with temperature.
Length: Resistance is directly proportional to length. A longer wire offers more resistance. $R \propto L$
Cross-sectional Area: Resistance is inversely proportional to the cross-sectional area. A thicker wire (larger area) offers less resistance. $R \propto \frac{1}{A}$
Temperature: For most metallic conductors, resistance increases with increasing temperature.

Conductivity ($\sigma$) is the reciprocal of resistivity ($\sigma = \frac{1}{\rho}$) and measures how easily current flows through a material. Materials with high conductivity have low resistivity and vice versa.

In this problem, the material (copper) and length are constant, so the difference in resistance is purely due to the difference in cross-sectional area resulting from the different radii.

Remembering how resistance scales with radius ($R \propto \frac{1}{r^2}$) is useful for quickly comparing resistances of wires made of the same material and length but different radii.

Prepp Team · Answer

2R

Prepp Team · Answer

$\frac{{\rm{R}}}{2}$

Prepp Team · Answer

$\frac{3}{4}{\rm{R}}$

Understanding Electrical Resistance in Wires

Analyzing the Two Copper Wires

Calculating the Resistance of Each Wire

Total Resistance in Series Combination

Revision Table: Key Concepts

Additional Information: Factors Affecting Resistance and Conductivity

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Concept	Description	Formula
Electrical Resistance (\(R\))	Opposition to current flow	\(R = \rho \frac{L}{A}\)
Resistivity (\(\rho\))	Material property indicating resistance	Units: Ohm-meter (\(\Omega \cdot m\))
Cross-sectional Area (\(A\))	Area of the conductor's face perpendicular to current flow	For wire: \(A = \pi r^2\)
Series Combination	Components connected end-to-end	\(R_{total} = R_1 + R_2 + ...\)