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Chapter 28: Problem 36

The Wheatstone bridge shown is being used to measure resistance \(X\). At balance, the current through the galvanometer \(\mathrm{G}\) is zero and resistances \(L, M\), and \(N\) are \(3.0 \Omega, 2.0\) \(\Omega\), and \(10 \Omega\), respectively. Find the value of \(X\).

Short Answer

Expert verified
The unknown resistance \( X \) is \( 6.67 \Omega \).

Step by step solution

01

Understanding the Wheatstone Bridge

A Wheatstone bridge is a circuit used to determine the unknown resistance by adjusting the other resistances until the current through the galvanometer is zero. At balance, the ratio of the resistances on one side of the bridge is equal to the ratio on the other side. This means that the ratio of \( L \) to \( M \) is equal to the ratio of \( N \) to \( X \).
02

Writing the Balance Equation

Since the bridge is balanced, we can write the equation for the ratio of resistors: \[ \frac{L}{M} = \frac{N}{X} \] where \( L = 3.0 \Omega \), \( M = 2.0 \Omega \), and \( N = 10 \Omega \). Substitute these values into the equation to find \( X \).
03

Solving for X

Replace \( L, M, \) and \( N \) with their values to get: \[ \frac{3.0}{2.0} = \frac{10}{X} \]. This simplifies to \( 1.5 = \frac{10}{X} \). Cross-multiply to solve for \( X \): \( 1.5X = 10 \). Divide both sides by 1.5 to get \( X = \frac{10}{1.5} \).
04

Calculating X

Perform the division: \( X = \frac{10}{1.5} = 6.67 \Omega \). Therefore, the unknown resistance \( X \) is \( 6.67 \Omega \).

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Key Concepts

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

Resistance Measurement
Resistance measurement is a crucial part of electrical engineering and physics. Understanding the resistance of a component helps you determine how much it resists the flow of electric current. The Wheatstone bridge is one of the most precise methods for measuring resistance.

The bridge consists of four resistances arranged in a diamond shape. The goal is to find the unknown resistance, labeled as \(X\), by ensuring no current flows through a connected galvanometer. This indicates that the bridge is balanced and the unknown resistance can be calculated.

In a balanced Wheatstone bridge, the ratio of the two known resistances in one branch is equal to the ratio of the known resistance and the unknown resistance in the other branch. By understanding this balance equation, we can measure the unknown resistance accurately.
Circuit Analysis
Circuit analysis is the study of how electrical components interact in a circuit. With a Wheatstone bridge, circuit analysis involves understanding how resistors work together to balance the bridge. Balancing occurs when the ratios of resistances on opposite sides of the bridge are equal. At this point, no current flows through the galvanometer, indicating equilibrium.

To analyze the Wheatstone bridge circuit, you write the balance equation. This equation relates the resistances \(L\), \(M\), \(N\), and \(X\) in the form \(\frac{L}{M} = \frac{N}{X}\). Analyzing this equation lets you solve for the unknown resistance. This method not only teaches fundamental concepts of circuits but also provides a practical tool for measuring resistance.
Galvanometer
A galvanometer is a sensitive instrument used to detect small electric currents. In a Wheatstone bridge, it's placed in the diagonal between the two branches. The goal is to achieve a zero reading on the galvanometer, indicating that the bridge is balanced.

The galvanometer helps identify when the voltage across it is zero, meaning there is no current flow. This occurs when the product of the two known resistors and their opposite counterparts in the bridge are equal. When balanced, the galvanometer effectively serves as an indicator of a successful resistance measurement.

By adjusting the known resistances, you can achieve this zero current. The galvanometer's role is thus critical in fine-tuning the circuit to determine the unknown resistance accurately.

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Most popular questions from this chapter

What resistance must be placed in parallel with \(20 \Omega\) to make the combined resistance \(15 \Omega ?\)

A \(120-\mathrm{V}\) house circuit has the following light bulbs turned on: \(40.0\) \(\mathrm{W}, 60.0 \mathrm{~W}\), and \(175.0 \mathrm{~W}\). Find the equivalent resistance of these lights. House circuits are so constructed that each device is connected in parallel with the others. From \(\mathrm{P}=V I=V^{2} / R\), for the first bulb $$ R_{1}=\frac{V^{2}}{P_{1}}=\frac{(120 \mathrm{~V})^{2}}{40.0 \mathrm{~W}}=360 \Omega $$ Similarly, \(R_{2}=240 \Omega\) and \(R_{3}=192 \Omega\). Because devices in a house circuit are in parallel, $$ \frac{1}{R_{\mathrm{eq}}}=\frac{1}{360 \Omega}+\frac{1}{240 \Omega}+\frac{1}{192 \Omega} \quad \text { or } \quad R_{\mathrm{eq}}=82.3 \Omega $$ As a check, note that the total power drawn from the line is \(40.0\) \(\mathrm{W}+60.0 \mathrm{~W}+75.0 \mathrm{~W}=75.0 \mathrm{~W}\). Then, using \(\mathrm{P}=V^{2} / R\) $$ R_{\mathrm{eq}}=\frac{V^{2}}{\text { total power }}=\frac{(120 \mathrm{~V})^{2}}{175.0 \mathrm{~W}}=82.3 \Omega $$

Compute the equivalent resistance of \(4.0 \Omega\) and \(8.0 \Omega(a)\) in series and \((b)\) in parallel.

By use of one or more of the three resistors \(3.0 \Omega, 5.0 \Omega\), and \(6.0\) \(\Omega\), a total of 18 resistances can be obtained. What are they?

Show that if two resistors are connected in parallel, the rates at which they produce thermal energy vary inversely as their resistances.
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