A copper wire of radius r and length l has a resistance of R. A second copper wire with radius 2r and length l is taken, and the two wires are joined in a parallel combination. The resultant resistance of the parallel combination of the two wires will be
R / 5
Understanding Electrical Resistance of a Wire
The electrical resistance of a wire depends on its material, length, and cross-sectional area. The formula for resistance (\(R\)) is given by:
\[R = \rho \frac{l}{A}\]
Where:
- \(\rho\) (rho) is the resistivity of the material (a property specific to the material, like copper).
- \(l\) is the length of the wire.
- \(A\) is the cross-sectional area of the wire.
For a circular wire with radius \(r\), the cross-sectional area is \(A = \pi r^2\).
Calculating Resistance of the Second Copper Wire
We are given that the first copper wire has a radius \(r\), length \(l\), and resistance \(R\). Using the formula:
\[R = \rho \frac{l}{\pi r^2} \quad (Equation \, 1)\]
The second copper wire is made of the same material (copper, so \(\rho\) is the same). It has a radius of \(2r\) and a length of \(l\). Let's call its resistance \(R_2\).
The radius of the second wire is \(r_2 = 2r\).
The cross-sectional area of the second wire is \(A_2 = \pi r_2^2 = \pi (2r)^2 = \pi (4r^2) = 4\pi r^2\).
Now, let's calculate the resistance \(R_2\) of the second wire:
\[R_2 = \rho \frac{l}{A_2} = \rho \frac{l}{4\pi r^2}\]
We can rewrite this expression by factoring out terms from Equation 1:
\[R_2 = \frac{1}{4} \left( \rho \frac{l}{\pi r^2} \right)\]
From Equation 1, we know that \(R = \rho \frac{l}{\pi r^2}\). So, we can substitute \(R\) into the expression for \(R_2\):
\[R_2 = \frac{1}{4} R\]
Thus, the resistance of the second copper wire is \(R/4\).
Calculating Resultant Resistance in Parallel Combination
The two wires (the first wire with resistance \(R_1 = R\) and the second wire with resistance \(R_2 = R/4\)) are joined in a parallel combination. The formula for the equivalent resistance (\(R_{parallel}\)) of two resistors connected in parallel is:
\[\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}\]
Substitute the values \(R_1 = R\) and \(R_2 = R/4\) into the formula:
\[\frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R/4}\]
To simplify the term \(\frac{1}{R/4}\), we can invert the denominator:
\[\frac{1}{R/4} = 1 \times \frac{4}{R} = \frac{4}{R}\]
Now, substitute this back into the parallel combination formula:
\[\frac{1}{R_{parallel}} = \frac{1}{R} + \frac{4}{R}\]
Combine the fractions on the right side:
\[\frac{1}{R_{parallel}} = \frac{1 + 4}{R} = \frac{5}{R}\]
To find \(R_{parallel}\), take the reciprocal of both sides:
\[R_{parallel} = \frac{R}{5}\]
Therefore, the resultant resistance of the parallel combination of the two wires is \(R/5\).
Step-by-Step Solution Summary
Here is a summary of the steps taken to find the resultant resistance:
- Identify the resistance of the first wire, \(R_1 = R\).
- Calculate the resistance of the second wire (\(R_2\)) based on its dimensions and material. Using the resistance formula \(R = \rho \frac{l}{A}\), we found \(R_2 = R/4\).
- Apply the formula for equivalent resistance of two resistors in parallel: \(\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}\).
- Substitute the values of \(R_1\) and \(R_2\) into the parallel formula and solve for \(R_{parallel}\).
The calculation yielded \(R_{parallel} = R/5\).
| Wire | Radius | Length | Area (\(A = \pi r^2\)) | Resistance (\(R = \rho l / A\)) |
|---|---|---|---|---|
| First Wire | \(r\) | \(l\) | \(\pi r^2\) | \(R_1 = R = \rho \frac{l}{\pi r^2}\) |
| Second Wire | \(2r\) | \(l\) | \(\pi (2r)^2 = 4\pi r^2\) | \(R_2 = \rho \frac{l}{4\pi r^2} = \frac{1}{4} \left(\rho \frac{l}{\pi r^2}\right) = \frac{R}{4}\) |
Revision Table: Electrical Resistance Concepts
| Concept | Description | Formula |
|---|---|---|
| Resistance (\(R\)) | Opposition to electric current flow. | \(R = \rho \frac{l}{A}\) |
| Resistivity (\(\rho\)) | Intrinsic property of a material indicating how strongly it resists current. | Unit: Ohm-meter (\(\Omega \cdot m\)) |
| Parallel Resistance | Equivalent resistance of components connected across the same potential difference. | \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...\) |
| Series Resistance | Equivalent resistance of components connected end-to-end, so current flows sequentially. | \(R_{eq} = R_1 + R_2 + ...\) |
Additional Information: Factors Affecting Resistance
Beyond the dimensions and material, the resistance of a conductor can also be affected by other factors, particularly temperature.
- Temperature: For most metallic conductors, resistance increases with increasing temperature. This is because higher temperatures cause atoms in the material to vibrate more vigorously, making it harder for electrons to flow through. The relationship is often approximately linear over a limited temperature range.
- Stress/Strain: Applying mechanical stress or strain to a wire can change its dimensions (length and area) and sometimes its resistivity, thereby changing its resistance. This principle is used in strain gauges.
Understanding how these factors influence resistance is crucial for designing electrical circuits and components accurately.
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