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Chapter 27: Problem 19

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?

Short Answer

Expert verified
The unknown resistance is 4.00 惟 and should be connected in parallel.

Step by step solution

01

Understanding the Problem

We need to find an unknown resistance, let's call it \( R_x \), that when combined with a known resistance of \( 12.0 \Omega \) gives a total resistance of \( 3.00 \Omega \). We should determine if they should be connected in series or parallel.
02

Identifying Combination Method

In series, the total resistance \( R_T \) is the sum of individual resistances: \( R_T = R_1 + R_2 \). In parallel, the formula is \( 1/R_T = 1/R_1 + 1/R_2 \). We aim for a total resistance lower than \( 12.0 \Omega \), indicating a parallel connection should be used.
03

Using the Parallel Formula

Since the resistors should be in parallel, we use the equation \( 1/R_T = 1/R_1 + 1/R_2 \). Plug in \( R_T = 3.00 \Omega \) and \( R_1 = 12.0 \Omega \), the formula becomes: \( 1/3.00 = 1/12.0 + 1/R_x \).
04

Solving for the Unknown Resistance

Rearrange the parallel equation to solve for \( 1/R_x \): \[ 1/R_x = 1/3.00 - 1/12.0 \]. Compute \( 1/3.00 = 0.3333 \) and \( 1/12.0 = 0.0833 \). Thus, \( 1/R_x = 0.3333 - 0.0833 = 0.250 \).
05

Calculating the Value of Unknown Resistance

Now take the reciprocal to find \( R_x \): \( R_x = 1/0.250 = 4.00 \Omega \).

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Key Concepts

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

Series and parallel circuits
Electric circuits often involve combining multiple resistors to achieve a desired resistance. Resistors can be connected in two primary ways: in series or in parallel.
  • Series connection: In a series circuit, resistors are connected end-to-end, one after the other. The total resistance is simply the sum of the individual resistances: \( R_T = R_1 + R_2 + ... + R_n \). This configuration results in a higher total resistance.
  • Parallel connection: In a parallel circuit, resistors are connected side by side. This arrangement allows multiple paths for current to flow. The formula for total resistance in parallel is \( 1/R_T = 1/R_1 + 1/R_2 + ... + 1/R_n \). This typically results in a lower total resistance than any individual resistance.
Understanding these two configurations is crucial when designing circuits to achieve specific resistance values. In our case, we need a total resistance that is less than any of the connected resistors, which leads us to use a parallel connection.
Ohm's law
Ohm's Law is a fundamental principle used in electric circuits to relate voltage, current, and resistance. It is expressed with the formula \( V = IR \), where:
  • \( V \): Voltage - the electrical potential difference measured in volts (V).
  • \( I \): Current - the flow of electric charge measured in amperes (A).
  • \( R \): Resistance - the opposition to the flow of current, measured in ohms (\( \Omega \)).
Ohm's Law helps in understanding how current flows through an electric circuit. By rearranging the formula, we can find any of the three variables if the other two are known. Although it was not directly used to solve the specific exercise, understanding Ohm's Law is fundamental to electric circuit analysis and helps in predicting how different circuit setups can affect voltage, current, and resistance.
Electric circuits analysis
Electric circuit analysis involves understanding how various components in a circuit interact and how they affect the overall function of the circuit. Here鈥檚 how you can approach such an analysis:
  • Identify all components: Recognize resistors, capacitors, power sources, etc., and note their values and connections.
  • Determine the circuit configuration: Assess if components are arranged in series, parallel, or a combination of both. This aids in using the correct formulas for analysis.
  • Apply Ohm鈥檚 Law and other relevant formulas: Once the arrangement is known, use appropriate formulas to calculate unknown values such as resistance, current, or voltage.
  • Solve for unknowns: Use algebraic methods to manipulate equations, finding the desired values like the unknown resistance in our case study.
In our exercise, analyzing the circuit helped us determine that a parallel connection was necessary to achieve the desired resistance. The correct analysis led to using the parallel resistance formula and solving for the unknown resistor value, illustrating practical application of these techniques.

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Most popular questions from this chapter

Two identical batteries of emf \(\mathscr{h}=\) \(12.0 \mathrm{~V}\) and internal resistance \(r=0.200 \Omega\) are to be connected to an external resistance \(R\), either in parallel (Fig. \(27-50\) ) or in series (Fig. 27-51). If \(R=2.00 r\), what is the current \(i\) in the external resistance in the (a) parallel and (b) series arrangements? (c) For which arrangement is \(i\) greater? If \(R=\) \(r / 2.00\), what is \(i\) in the external resistance in the (d) parallel arrangement and (e) series arrangement? (f) For which arrangement is \(i\) greater now?

A solar cell generates a potential difference of \(0.10 \mathrm{~V}\) when a \(500 \Omega\) resistor is connected across it, and a potential difference of \(0.15 \mathrm{~V}\) when a \(1000 \Omega\) resistor is substituted. What are the (a) internal resistance and (b) emf of the solar cell? (c) The area of the cell is \(5.0 \mathrm{~cm}^{2}\), and the rate per unit area at which it receives energy from light is \(2.0 \mathrm{~mW} / \mathrm{cm}^{2}\). What is the efficiency of the cell for converting light energy to thermal energy in the \(1000 \Omega\) external resistor?

(a) In electron-volts, how much work does an ideal battery with a \(12.0 \mathrm{~V}\) emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If \(3.40 \times 10^{18}\) electrons pass through each second, what is the power of the battery in watts?

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a \(0.220 \mu \mathrm{F}\) capacitor. The capacitor is charged to \(5.00 \mathrm{~V}\), then discharged through the resistor. The time for the potential difference across the plates to decrease to \(0.800 \mathrm{~V}\) is measured by a clock inside the game. If the range of discharge times that can be handled effectively is from \(10.0 \mu \mathrm{s}\) to \(6.00 \mathrm{~ms}\), what should be the (a) lower value and (b) higher value of the resistance range of the resistor?

Side flash. Figure \(27-38\) indicates one reason no one should stand under a tree during a lightning storm. If lightning comes down the side of the tree, a portion can jump over to the person, especially if the current on the tree reaches a dry region on the bark and thereafter must travel through air to reach the ground. In the figure, part of the lightning jumps through distance \(d\) in air and then travels through the person (who has negligible resistance relative to that of air because of the highly conducting salty fluids within the body). The rest of the current travels through air alongside the tree, for a distance \(h\). If \(d / h=0.400\) and the total current is \(I=5000 \mathrm{~A}\), what is the current through the person?
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