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Chapter 27: Problem 86

Two resistors \(R_{1}\) and \(R_{2}\) may be connected either in series or in parallel across an ideal battery with emf \(\mathscr{E}\). We desire the rate of energy dissipation of the parallel combination to be five times that of the series combination. If \(R_{1}=100 \Omega\), what are the (a) smaller and (b) larger of the two values of \(R_{2}\) that result in that dissipation rate?

Short Answer

Expert verified
The possible values of \(R_2\) are 50 \(\Omega\) and 400 \(\Omega\).

Step by step solution

01

Understand the power in series

When two resistors are connected in series, the total resistance is the sum: \(R_{\text{series}} = R_{1} + R_{2}\). The power dissipated in a series circuit, given an emf \(\mathscr{E}\) and series resistance, is \(P_{\text{series}} = \frac{\mathscr{E}^2}{R_{\text{series}}}\).
02

Understand the power in parallel

For two resistors in parallel, the total resistance is given by the formula \(\frac{1}{R_{\text{parallel}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\). The power dissipated in a parallel circuit is \(P_{\text{parallel}} = \frac{\mathscr{E}^2}{R_{\text{parallel}}}\).
03

Relate power dissipation

According to the problem, the power dissipation in the parallel circuit is five times that in the series circuit: \(P_{\text{parallel}} = 5 \times P_{\text{series}}\). Thus, the equation becomes \(\frac{\mathscr{E}^2}{R_{\text{parallel}}} = 5 \times \frac{\mathscr{E}^2}{R_{\text{series}}}\).
04

Set up the equation

Using the relation from Step 3, cancel \(\mathscr{E}^2\) and set up the equation: \(\frac{1}{R_{\text{parallel}}} = 5 \times \frac{1}{R_{\text{series}}}\). Substitute the expressions for \(R_{\text{parallel}}\) and \(R_{\text{series}}\).
05

Substitute values and simplify

Substitute \(R_{1} = 100 \Omega\) in the equation and solve: \(\frac{1}{R_{\text{parallel}}} = 5 \times \frac{1}{100 + R_{2}}\). Using \(\frac{1}{R_{\text{parallel}}} = \frac{1}{100} + \frac{1}{R_{2}}\), set up the equation as: \(\frac{1}{100} + \frac{1}{R_{2}} = \frac{5}{100 + R_{2}}\).
06

Solve the equation for \(R_{2}\)

Rearrange to form a quadratic equation: \(5R_{2}(100 + R_{2}) = 100R_{2} + 100^2\). Expand and simplify: \(5R_{2}^2 + 500R_{2} = 100R_{2} + 10000\). Rearrange to form: \(5R_{2}^2 + 400R_{2} - 10000 = 0\).
07

Solve the quadratic equation for \(R_{2}\)

Use the quadratic formula \(R_{2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = 400\), and \(c = -10000\). Calculate the discriminant \(b^2 - 4ac = 400^2 - 4(5)(-10000)\) and proceed with finding the roots.
08

Calculate the values of \(R_{2}\)

Solve \(R_{2} = \frac{-400 \pm \sqrt{400^2 + 200000}}{10}\). Simplifying, compute \(R_{2} = \frac{-400 \pm \sqrt{360000}}{10}\). The square root yields two possible values of \(R_{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series and Parallel Resistors
Resistors can be combined in electric circuits in two common configurations: series and parallel. When resistors are connected in series, their resistance values add up. For example, if you have two resistors, R鈧� and R鈧�, in series, the total resistance is defined as:
  • \( R_{\text{series}} = R_{1} + R_{2} \)
The resistance increases because the electric current has to pass through both resistors sequentially.
In contrast, when resistors are connected in parallel, the total resistance is lower than any individual resistor. This is due to multiple pathways available for the current to pass. The total resistance in parallel, for two resistors, is calculated using:
  • \( \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \)
Understanding how to calculate these equivalent resistances is critical for analyzing complex circuits.
Power Dissipation
Power dissipation in a circuit refers to how much energy is converted to heat or work. It occurs any time electrical energy is used in a component like a resistor. Power (\( P \)) in terms of a resistor and a voltage source is calculated by the formula:
  • \( P = \frac{\mathscr{E}^2}{R} \)
where \( \mathscr{E} \) is the electromotive force (emf) or voltage, and \( R \) is the resistance.
In the context of the exercise, we have two situations: one with resistors in series and another with resistors in parallel. It's crucial to know that the power dissipation can vary significantly between these configurations. For example, parallel circuits often dissipate more power because the total resistance is less, leading to increased current flow.
Quadratic Equation
A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \). Solving it requires determination of its roots. In many circuit problems similar to the one given, quadratic equations appear due to the relationships between resistances and other parameters.
The solution to a quadratic equation is found using the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In the exercise, after substituting specific values for resistors and simplifying, we use this formula to find the values of \( R_{2} \). Here, \( a = 5 \), \( b = 400 \), and \( c = -10000 \). Ensure to correctly calculate the discriminant \( b^2 - 4ac \) to find the two possible solutions for the resistance.
Ohm's Law
Ohm's Law is a fundamental principle in electronics, connecting voltage \( V \), current \( I \), and resistance \( R \) in electrical circuits. The law is expressed as:
  • \( V = IR \)
This means that the voltage across a circuit is the product of the current flowing through it and its resistance.
In the exercise, while not stated directly, understanding Ohm's Law helps deduce how changes in resistance affect current and voltage. Knowing this relationship provides insight into why the power dissipation differs between series and parallel resistors. In series circuits, the total resistance limits current flow more than parallel circuits do, thus impacting the total subsystem's power dissipation. Using Ohm's Law in conjunction with other formulas is essential in solving these circuit problems effectively.

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Most popular questions from this chapter

The starting motor of a car is turning too slowly, and the mechanic has to decide whether to replace the motor, the cable, or the battery. The car's manual says that the \(12 \mathrm{~V}\) battery should have no more than \(0.020 \Omega\) internal resistance, the motor no more than \(0.200 \Omega\) resistance, and the cable no more than \(0.040 \Omega\) resistance. The mechanic turns on the motor and measures \(11.4 \mathrm{~V}\) across the battery, \(3.0 \mathrm{~V}\) across the cable, and a current of \(50 \mathrm{~A}\). Which part is defective?

A capacitor with initial charge \(q_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

A capacitor with an initial potential difference of \(100 \mathrm{~V}\) is discharged through a resistor when a switch between them is closed at \(t=0 .\) At \(t=10.0 \mathrm{~s}\), the potential difference across the capacitor is \(1.00 \mathrm{~V}\). (a) What is the time constant of the circuit? (b) What is the potential difference across the capacitor at \(t=17.0 \mathrm{~s}\) ?

A total resistance of \(3.00 \Omega\) is to be produced by connecting an unknown resistance to a \(12.0 \Omega\) resistance. (a) What must be the value of the unknown resistance, and (b) should it be connected in series or in parallel?

The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other) \(2.0 \mu \mathrm{F}\) capacitor drops to one-fourth its initial value in \(2.0 \mathrm{~s}\). What is the equivalent resistance between the capacitor plates?
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