Problem 28 - Technical Mathematics with Calculus

Question

Referring to Fig. \(12-9,\) (a) determine what two resistances will give a total resistance of \(780 \Omega\) when wired in series and \(105 \Omega\) when wired in parallel. (b) Find two resistances that will give an equivalent resistance of \(9070 \Omega\) in series and \(1070 \Omega\) in parallel. (Recall that the equivalent resistance for two resistors \(R_{1}\) and \(R_{2}\) is \(R_{1}+R_{2}\) when wired in series, and \(R_{1} R_{2} /\left(R_{1}+R_{2}\right)\) when wired in parallel.)

Step-by-Step Solution

Answer

For part (a), solve the quadratic equation from step 4 to find the exact values for R1 and R2. Repeat the process for part (b) using the quadratic equation from step 8.

Step 1: Understand the relationship between resistances in series and parallel

When two resistances, R1 and R2, are connected in series, their total resistance R_series is given by R_series = R1 + R2. In parallel, their total resistance R_parallel is the product over the sum: R_parallel = (R1 * R2) / (R1 + R2).

Step 2: Set up equations for part (a)

For the given total resistance (780 Ohm in series and 105 Ohm in parallel), we set up two equations: R1 + R2 = 780 and (R1 * R2) / (R1 +R2) = 105.

Step 3: Solve for one resistance in terms of the other for part (a)

From the parallel resistance equation (R1 * R2) / (R1 + R2) = 105, express R2 in terms of R1: R2 = (105 * R1) / (R1 - 105).

Step 4: Substitute and solve the quadratic equation for part (a)

Substitute R2 into the series resistance equation: R1 + (105 * R1) / (R1 - 105) = 780. This leads to a quadratic equation in R1 that can be solved.

Step 5: Find the resistances for part (a)

Solve the quadratic equation for R1 and use the result to calculate R2 using the expression from the previous step. These values are the two resistances.

Step 6: Set up equations for part (b)

The total resistance is now 9070 Ohm in series and 1070 Ohm in parallel. The equations are R1 + R2 = 9070 and (R1 * R2) / (R1 + R2) = 1070.

Step 7: Solve for one resistance in terms of the other for part (b)

Use the parallel resistance formula to express R2 in terms of R1 for part (b): R2 = (1070 * R1) / (R1 - 1070).

Step 8: Substitute and solve the quadratic equation for part (b)

Substitute this expression for R2 into the series resistance equation: R1 + (1070 * R1) / (R1 - 1070) = 9070. This yields another quadratic equation.

Step 9: Find the resistances for part (b)

Solve the quadratic equation for R1 and then find R2 using the formula derived in the previous step for part (b).

Key Concepts

Series and Parallel Circuits

Understanding series and parallel circuits is fundamental in electronics and electrical engineering. These concepts refer to the two primary ways that components like resistors can be connected within a circuit.

Series circuits are those in which components are connected end-to-end, forming a single path for the current to flow. The total resistance in a series circuit is simply the sum of all the resistors. On the other hand, parallel circuits have branches, each of which is a separate path for the current to flow. The total resistance in a parallel circuit is a bit more complex to calculate, as it takes into account the resistance of each branch and how they influence one another.

Understanding these configurations is crucial when we calculate total resistance in a circuit, or when we want to design a circuit to have a specific resistance. These principles also apply to other components, such as capacitors and inductors, which behave differently when arranged in series or parallel.

Resistors in Series

When discussing resistors in series, think of them as being lined up in a single file. The current that flows through one resistor must flow through the next one, all the way down the line. In this way, resistors add their resistances together, and the total resistance is simply the sum.

For example, if you have two resistors, say \( R_1 \) and \( R_2 \) connected in series, the total resistance \( R_{series} \) can be calculated with the formula:

• \( R_{series} = R_1 + R_2 \)

This is because the same current must pass consecutively through \( R_1 \) and \( R_2 \) causing their resistances to add up directly.

Resistors in Parallel

When we place resistors in parallel, each resistor is connected to the same two points, creating multiple pathways for the current. The total resistance in a parallel circuit results in a value that is less than the smallest individual resistor in the parallel network.

The formula to find the total resistance \( R_{parallel} \) for two resistors, \( R_1 \) and \( R_2 \) in parallel is:

• \( R_{parallel} = \frac{R_1 * R_2}{R_1 + R_2} \)

This equation can seem counterintuitive at first because adding more resistors reduces the total resistance. However, by providing more paths for the current to take, the circuit allows more current to flow, which reduces the overall resistance.

Quadratic Equations

Let's delve into quadratic equations, which are polynomial equations of degree two. The general form of a quadratic equation is:

• \( ax^2 + bx + c = 0 \)

where \( a, b, \) and \( c \) are constants, and \( a \) is not equal to zero.

These equations are pivotal in many scientific and engineering calculations, including finding the values of resistors in complex circuits. To solve a quadratic equation, we can use factoring, completing the square, or the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In the context of calculating resistances for circuits, forming and solving quadratic equations help us determine the specific resistor values that will satisfy certain conditions in both series and parallel configurations.

Learning to solve quadratic equations can open up a range of possibilities in analyzing and designing circuits, as seen in our exercise involving the determination of resistor values based on specified total resistances in series and parallel connections.