Problem 20 - Technical Mathematics with Calculus

Question

The equivalent resistance of two resistors in parallel is $$\frac{R_{1} R_{2}}{R_{1}+R_{2}}$$ If each resistor is made of wire of resistivity \(\rho,\) with \(R_{1}\) using a wire of length \(L_{1}\) and cross-sectional area \(A_{1},\) and \(R_{2}\) having a length \(L_{2}\) and area \(A_{2}\) our expression becomes $$\frac{\frac{\rho L_{1}}{A_{1}} \cdot \frac{\rho L_{2}}{A_{2}}}{\frac{\rho L_{1}}{A_{1}}+\frac{\rho L_{2}}{A_{2}}}$$ Simplify this complex fraction.

Step-by-Step Solution

Answer

\( R_{eq} = \frac{\rho L_{1}L_{2}}{A_{1} + A_{2}} \)

Step 1: Establish Variables for Each Resistor's Resistance

The resistance of a wire is given by the formula: \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire. For the two resistors in question: Resistor 1 has resistance \( R_{1} = \frac{\rho L_{1}}{A_{1}} \) and Resistor 2 has resistance \( R_{2} = \frac{\rho L_{2}}{A_{2}} \).

Step 2: Substitute the Expressions for \( R_{1} \) and \( R_{2} \)

Substitute the calculated values of \( R_{1} \) and \( R_{2} \) into the formula for equivalent resistance in a parallel circuit: \( R_{eq} = \frac{R_{1}R_{2}}{R_{1} + R_{2}} \). This gives us: \( R_{eq} = \frac{\frac{\rho L_{1}}{A_{1}} \cdot \frac{\rho L_{2}}{A_{2}}}{\frac{\rho L_{1}}{A_{1}} + \frac{\rho L_{2}}{A_{2}}} \).

Step 3: Factor Out the Resistivity \( \rho \)

Factor out the \( \rho \) term, which is common to both the numerator and denominator: \( R_{eq} = \frac{\rho^2 L_{1}L_{2}}{A_{1}A_{2}} \cdot \frac{1}{\frac{\rho L_{1}}{A_{1}} + \frac{\rho L_{2}}{A_{2}}} \).

Step 4: Combine the Terms in the Denominator

Combine the terms in the denominator by finding a common denominator: \( \frac{A_{2}}{A_{1}A_{2}} + \frac{A_{1}}{A_{1}A_{2}} \), which simplifies to \( \frac{A_{1} + A_{2}}{A_{1}A_{2}} \).

Step 5: Simplify the Equation

The equation now reads: \( R_{eq} = \frac{\rho^2 L_{1}L_{2}}{A_{1}A_{2}} \cdot \frac{A_{1}A_{2}}{\rho (A_{1} + A_{2})} \). Cancel out the \( \rho \) from the numerator and denominator, as well as the \( A_{1}A_{2} \) terms, to simplify the formula to \( R_{eq} = \frac{\rho L_{1}L_{2}}{A_{1} + A_{2}} \).

Key Concepts

Resistivity

Resistivity is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. Defined as the electrical resistance of a conductor with unit length and unit cross-sectional area, it is expressed using the Greek letter \( \rho \). The resistivity of a material depends on its nature and temperature, affecting how it conducts electricity. For instance, metals typically have low resistivities, meaning they conduct electricity well.

In many physics problems, we use resistivity to calculate the resistance of a wire or a component made from a specific material. As seen in our problem, the resistance \( R \) of each resistor is determined by its material's resistivity \( \rho \) in combination with its length \( L \) and cross-sectional area \( A \). So, in an equation, we express the resistance as \( R = \frac{\rho L}{A} \), signifying that a longer wire or one with a smaller cross-sectional area will have higher resistance. Conversely, materials with lower resistivity, or larger cross-sectional areas, result in lower resistance.

Electrical Resistance

Electrical resistance, symbolized by \( R \), measures an object's opposition to the flow of electric current. It's analogous to the friction experienced by water when flowing through a pipe; the higher the resistance, the harder it is for current to flow. Resistance is measured in ohms (\(\Omega\)), and it's not just dependent on the material's properties but also on its dimensions.

Through Ohm's Law \( V = IR \), which relates voltage \( V \) to current \( I \) and resistance \( R \) in a linear relationship, we see how resistance affects electrical circuits. This formula imparts a fundamental understanding that for a constant voltage, increased resistance will lead to a decrease in current. In our exercise, calculating the total resistance of resistors in a circuit is crucial for understanding how they influence the overall electric current that can flow through the circuit.

Parallel Circuits

Parallel circuits provide multiple paths for electricity to flow. This configuration is fundamental in electronic appliances and systems, as it ensures that if one pathway fails, the circuit can still operate via other paths. The unique feature of a parallel circuit is that it maintains the same voltage across all components connected in parallel.

When it comes to calculating the equivalent resistance of parallel resistors, the formula we use is different compared to series circuits. The equivalent resistance \( R_{eq} \) of resistors in parallel is always lower than the smallest resistor in the circuit and is found by using the expression \( R_{eq} = \frac{R_{1}R_{2}}{R_{1} + R_{2}} \) for two resistors. This results in a decrease in the total resistance and an increase in the total current that can flow through the circuit. The solution for our exercise involves evaluating this formula, which ultimately enables us to understand how different resistor combinations affect the circuit's overall resistance.

Simplification of Complex Fractions

Complex fractions, which are fractions where the numerator, denominator, or both contain fractions themselves, are common in advanced math and physics problems. Simplifying complex fractions makes them easier to work with and understand. One standard method involves finding a common denominator in the numerator and the denominator, multiplying both parts of the fraction by this common denominator, and then reducing the resultant fraction.

In the given problem, the original expression for the equivalent resistance in a parallel circuit is a complex fraction. To simplify it, common factors such as resistivity \( \rho \) and cross-sectional areas \( A_{1} \) and \( A_{2} \) in the numerator and the denominator are first identified and then cancelled out where possible. This process, which is a practical application of simplifying complex fractions, leads to a more manageable expression that can be easily interpreted and used for further calculations or analyses within electrical circuit problems.