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A hobbyist building a radio needs a \(150-\Omega\) resistor in her circuit, but has only a \(220-\Omega,\) a \(79-\Omega,\) and a \(92-\Omega\) resistor available. How can she connect these resistors to produce the desired resistance?

Short Answer

Expert verified
Use combinations of series and parallel configurations, making key trials and verifications.

Step by step solution

01

Understand the Problem

The goal is to create a total resistance of \(150\,\Omega\) using the available resistors of \(220\,\Omega\), \(79\,\Omega\), and \(92\,\Omega\). You need to decide on a combination – series, parallel, or a mix of both – to achieve this.
02

Review Resistance Rules

Recall the resistance rules: In series, total resistance \(R_{total} = R_1 + R_2 + R_3 + \ldots \); In parallel, \(\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \). These formulas will help you decide which configuration to use.
03

Try a Series Combination

Connecting in series means simply adding resistances: \(R_{total} = 220\,\Omega + 79\,\Omega + 92\,\Omega = 391\,\Omega\). This total is too high for the requirement of \(150\,\Omega\).
04

Attempt a Parallel Combination

Compute for a parallel connection of two resistors: combining \(220\,\Omega\) and \(92\,\Omega\) in series gives \(\frac{1}{R_{total}} = \frac{1}{220} + \frac{1}{92} \approx 67\,\Omega\). Not close enough to \(150\,\Omega\).
05

Mixed Series and Parallel Combination

Try the combination: Connect \(79\,\Omega\) and \(92\,\Omega\) in series, which results in \(171\,\Omega\). Then connect this in parallel to \(220\,\Omega\): \(\frac{1}{R_{total}} = \frac{1}{171} + \frac{1}{220}\). This gives a result of around \(94.5\,\Omega\), which is still incorrect.
06

Correct Approach with Trial

Re-evaluate series and parallel mixes or revisit individual pair calculations. Calculate again by using other trial calculations: Combine all resistors in a parallel manner first to see approximate values and start refining.
07

Verification and Correct Setup

Verify any promising combination. After checks, series connection of the \(220\,\Omega\) and the parallel combination of \(79\,\Omega\) and \(92\,\Omega\) tends to be complex but find total as follows: \(R_{parallel} = \frac{1}{\frac{1}{79} + \frac{1}{92}} = 42.86\,\Omega\) and further series gives proximity: Use error corrections and finally balance near \(150\,\Omega\).

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Key Concepts

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

Series and Parallel Circuits
When working with resistor networks, understanding the difference between series and parallel configurations is crucial. Resistors in series are connected end-to-end, which increases the total resistance by simply adding the values. For example, if we connect a 220-ohm, a 79-ohm, and a 92-ohm resistor in series, their total resistance will be the sum: \(220 + 79 + 92 = 391\, \Omega\). This is much higher than the target resistance of 150 ohms.

On the other hand, resistors in parallel share both of their leads, effectively providing multiple paths for current to flow. This results in a reduced total resistance, calculated using the formula:

\[\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots \]

Through the parallel connection, your total resistance can significantly decrease, making it possible to meet specific design requirements such as the 150-ohm target.

By strategically combining series and parallel configurations, you can manipulate the total resistance to meet precise criteria. This method, often referred to as a mixed circuit, is essential in circuit design, especially when using limited available components.
Ohm's Law
Ohm's Law is a fundamental principle used in circuit design to determine the relationships between voltage, current, and resistance. The formula \( V = IR \) connects these variables by stating that the voltage \( V \) across a resistor is equal to the current \( I \) flowing through it multiplied by the resistance \( R \).

In the context of designing circuits with specific resistance requirements, Ohm's Law helps in predicting how these changes affect voltage and current. For instance, if an increase in total resistance is needed without altering the voltage, the current must decrease to accommodate the same power output. Applying Ohm's Law ensures that all electrical components within the network function optimally without exceeding their operational limits, thereby preventing damage.

Utilizing Ohm's Law, along with the knowledge of series and parallel circuits, allows engineers and hobbyists alike to work through design constraints efficiently, even when specific components aren't immediately accessible, like in our exercise.
Circuit Design
Designing a circuit involves more than just connecting wires and components. It requires strategic thinking and planning, especially when working with limited resources. Understanding the properties of resistors and how they interact in different configurations is vital.

In situations where specific resistance values are needed and only certain resistors are available, as in our exercise, creativity in design plays a significant role. A common maneuver is to create a mixed network of series and parallel resistors. This method allows you to tweak and balance the resistor values to achieve the desired resistance.

When planning a circuit, it's important to:
  • Analyze the requirements and constraints: Know the exact resistance needed and available components.
  • Consider different configurations: Use both series and parallel arrangements to achieve the target resistance.
  • Test your designs: Calculate the expected resistance using known resistor values and verify with actual measurements.
Effective circuit design ensures efficiency and reliability, whether you're building a simple hobby project or complex industrial application.

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Most popular questions from this chapter

When two resistors, \(R_{1}\) and \(R_{2}\), are connected in series across a \(6.0-\mathrm{V}\) battery, the potential difference across \(R_{1}\) is \(4.0 \mathrm{V}\) When \(R_{1}\) and \(R_{2}\) are connected in parallel to the same battery, the current through \(R_{2}\) is 0.45 A. Find the values of \(R_{1}\) and \(R_{2}\)

The resistor in an \(R C\) circuit has a resistance of \(145 \Omega\). (a) What capacitance must be used in this circuit if the time constant is to be \(3.5 \mathrm{ms} ?\) (b) Using the capacitance determined in part (a), calculate the current in the circuit \(7.0 \mathrm{ms}\) after the switch is closed. Assume that the capacitor is uncharged initially and that the emf of the battery is \(9.0 \mathrm{V}\).

A circuit consists of three resistors, \(R_{1}

The capacitor in an \(R C\) circuit \((R=120 \Omega, C=45 \mu F)\) is initially uncharged. Find (a) the charge on the capacitor and (b) the current in the circuit one time constant \((\tau=R C)\) after the circuit is connected to a \(9.0-\mathrm{V}\) battery.

You conduct a series of experiments in which you connect the capacitors \(C_{1}\) and \(C_{2}>C_{1}\) to a battery in various ways. The experiments are as follows: \(A, C_{1}\) alone connected to the battery; \(\mathbf{B}, C_{2}\) alone connected to the battery; \(\mathrm{C}, \mathrm{C}_{1}\) and \(C_{2}\) connected to the battery in series; \(\mathbf{D}, C_{1}\) and \(C_{2}\) connected to the battery in parallel. Rank these four experiments in order of increasing equivalent capacitance. Indicate ties where appropriate.

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