Question

Two wires of equal diameters of resistivity's \(\rho_{1}\) and \(\rho_{2}\) are joined in series. The equivalent resistivity of the combination is.... (A) \(\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}\) (B) \(\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}\) (C) \(\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}\) (D) \(\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}\)

Short answer

The equivalent resistivity of the combination of two wires of equal diameters and resistivities \(\rho_{1}\) and \(\rho_{2}\) joined in series is: \(\rho_{eq} = \frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{\ell_{1}+\ell_{2}}\).

Step-by-step solution

Expressing the resistance of each wire

The resistance of a wire is given by: \[R = \frac{\rho L}{A}\] Where \(R\) is resistance, \(\rho\) is resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. Here, we have two wires with equal diameters and resistivities \(\rho_{1}\) and \(\rho_{2}\), and lengths \(\ell_{1}\) and \(\ell_{2}\), with cross-sectional areas \(A_{1}\) and \(A_{2}\). Since the diameters are equal, their cross-sectional areas are also equal. Thus, we have \(A_{1} = A_{2}\).

Finding the equivalent resistance

If these two resistors are connected in series, their equivalent resistance is given by the sum of the individual resistances: \[R_{eq} = R_1 + R_2\]

Applying the resistance equation

Now, we apply the resistance formula for each wire, having \(R_1 = \frac{\rho_{1} \ell_{1}}{A_1}\) and \(R_2 = \frac{\rho_{2} \ell_{2}}{A_2}\): \[R_{eq} = \frac{\rho_{1} \ell_{1}}{A_1} + \frac{\rho_{2} \ell_{2}}{A_2}\] Since \(A_1 = A_2\), we can express the equivalent resistance as: \[R_{eq} = \frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{A}\]

Determining the equivalent resistivity

We know that the total length of the combination is \(\ell_{1}+\ell_{2}\). Similarly, we can express the equivalent resistance as the product of the equivalent resistivity, \(\rho_{eq}\), and total length of the combination, divided by the cross-sectional area, as: \[R_{eq} = \frac{\rho_{eq} (\ell_{1}+\ell_{2})}{A}\]

Comparing the equivalent resistance expressions

Comparing the expressions from Step 3 and Step 4: \[\frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{A} = \frac{\rho_{eq} (\ell_{1}+\ell_{2})}{A}\] By simplifying and isolating the equivalent resistivity \(\rho_{eq}\): \[\rho_{eq} = \frac{\rho_{1} \ell_{1} + \rho_{2} \ell_{2}}{\ell_{1}+\ell_{2}}\] Our answer corresponds to option (A) in the given choices.

Key concepts

Resistivity

Resistivity is a fundamental property of materials that measures their ability to resist the flow of electric current. It's often denoted by the symbol \(\rho\). Every material has its distinct resistivity value, which can depend on factors such as temperature and composition.

In the context of wires and electrical circuits, resistivity determines how much a given material will hinder the flow of electrons. It's worth noting that higher resistivity signifies that the material is more resistant to current flow, while a lower resistivity means it's easier for current to pass through.

The relationship between resistivity (\(\rho\)), resistance (\(R\)), length (\(L\)), and cross-sectional area (\(A\)) is given by the formula: \[ R = \frac{\rho L}{A} \] This crucial formula helps us calculate the resistance of a wire and understand how various factors impact electron flow in different materials.

Series Connection

When electrical components, like resistors or wires, are connected in series, they're aligned end-to-end in a single path for the current to travel. This means that the current has only one route, passing through each component sequentially.

A significant feature of a series connection is how it behaves with respect to resistance. For components connected in series, the total or equivalent resistance (\(R_{eq}\)) is simply the sum of their individual resistances: \[ R_{eq} = R_1 + R_2 + R_3 + \, ... \, + R_n \]
This cumulative property explains why a series of components shares the same electric current but may have different voltage drops across each element depending on their individual resistances. This collective resistance effect makes series connections advantageous for controlling current flow in circuits.

Resistance Formula

The resistance formula is key to determining how each component in an electrical circuit behaves. Resistance measures how strongly an object opposes the flow of an electric current. It is influenced by the material's resistivity, the length of the object, and its cross-sectional area.

The formula to calculate resistance is: \[ R = \frac{\rho L}{A} \]
This equation emphasizes that:- A longer wire increases resistance, as electrons have further to travel.- A larger cross-sectional area decreases resistance, as there is more space for electrons to move.- A higher resistivity material will increase resistance as it inherently opposes current flow more than a material with lower resistivity. Understanding this relationship aids in designing circuits with desired electrical properties by choosing suitable materials and geometries.

Cross-sectional Area

The cross-sectional area of a wire is crucial in determining its resistance. This area is a mathematical representation of the two-dimensional space perpendicular to the wire's length.

For a cylindrical wire, if the diameter is known, the cross-sectional area (\(A\)) can be calculated using the formula for the area of a circle: \[ A = \pi \left(\frac{d}{2}\right)^2 \] where \(d\) is the diameter.

The importance of cross-sectional area lies in its inverse relationship with resistance. With all other factors constant, a larger cross-sectional area means a lower resistance in the wire, allowing more electrons to pass through simultaneously. Hence, modifications to the cross-sectional area can be used to tailor the electrical properties of wire for specific applications, such as reducing heat buildup in circuits by using a wire with a larger area to minimize resistance.