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A relay having a resistance of \(6.0 \Omega\) operates with a minimum current of \(0.030 \mathrm{~A}\). It is required that the relay operate when the current in the line reaches \(0.240 \mathrm{~A}\). What resistance should be used to shunt the relay?

Short Answer

Expert verified
The shunt resistance should be approximately \(0.857 \Omega\).

Step by step solution

01

Understand the Problem

We have a relay with a resistance of \(6.0 \Omega\) that operates with a minimum current of \(0.030 \mathrm{~A}\). We need to find the resistance of a shunt to ensure the relay operates when the total line current is \(0.240 \mathrm{~A}\).
02

Define the Known Equations

Using Ohm's Law \(V = IR\), where \(I\) is current and \(R\) is resistance. Also, the concept of parallel resistances: when a shunt (parallel resistance) is added, the voltage across the relay must be equal to the voltage across the shunt.
03

Calculate Voltage across the Relay

The voltage required to operate the relay can be calculated using Ohm's Law: \( V = 0.030 \, \mathrm{A} \times 6.0 \, \Omega = 0.180 \, \mathrm{V} \). This is the voltage that must be maintained across the relay when the total line current is \(0.240 \, \mathrm{A}\).
04

Determine Total Current and Shunt Current

With a total current of \(0.240 \, \mathrm{A}\) in the line, the current through the shunt will be \(0.240 \, \mathrm{A} - 0.030 \, \mathrm{A} = 0.210 \, \mathrm{A}\).
05

Use Parallel Resistance Formula

For parallel circuits, the same voltage is across each component. Using the formula for parallel resistances, \( V_{shunt} = I_{shunt} \times R_{shunt} = 0.180 \, \mathrm{V} \), we solve for \( R_{shunt} \): \( R_{shunt} = \frac{0.180 \, \mathrm{V}}{0.210 \, \mathrm{A}}\).
06

Solve for Shunt Resistance

Calculate \( R_{shunt} \): \( R_{shunt} = \frac{0.180}{0.210} \approx 0.857 \Omega \). Thus, the resistance of the shunt should be approximately \(0.857 \Omega\).

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Key Concepts

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

Parallel Circuits
Parallel circuits are an essential topic when it comes to understanding electrical systems. In a parallel circuit, the electrical components are arranged in such a way that they each have their own separate loop to the power source. This means that the voltage across each component in a parallel circuit is the same.
Parallel circuits are useful in many practical applications because if one component fails, the current can still pass through the other components, allowing the system to function. This is different from a series circuit, where if one component fails, the entire circuit is broken.
  • Each component has the same voltage.
  • The total current is the sum of the currents through each component.
  • If one component fails, the current continues to flow through other paths.
This setup of parallel circuits applies directly to dealing with relay circuits, where maintaining consistent voltage across each component is crucial for operation.
Resistance Calculations
Calculating resistance in circuits, especially parallel circuits, is crucial for determining how the different pathways affect the overall electrical flow. When dealing with parallel resistors, the total resistance can be found using the formula: \[\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \]This formula signifies that the total resistance in a parallel circuit is always less than the smallest branch resistance. In the original exercise, it's important to find the resistance of a shunt resistor that will ensure the relay operates under specific conditions.
  • Use Ohm’s Law for calculating the voltage across components.
  • Remember that total parallel resistance is lower than any individual resistance.
  • The computation ensures the desired current flow through each path.
Understanding these calculations allow for precise configuration of circuit components to achieve desired electrical outcomes.
Electrical Current
Electrical current is the flow of electric charge and one of the fundamental concepts in electrical engineering. Current is measured in amperes (A) and represents how much charge is flowing through the circuit per unit of time. In both series and parallel circuits, calculating current accurately is vital for analyzing and designing electrical systems.
In the context of the original problem, the goal was to determine how much current would flow through the shunt given different conditions in the circuit. This is achieved by recognizing that in a parallel configuration, each component receives the same voltage, and the total current is divided.
  • Current is measured in amperes (A).
  • Ohm’s Law relates current, voltage, and resistance.
  • In parallel circuits, the total current is the sum of currents through each pathway.
Understanding electrical current helps determine the needed components for safe and efficient circuit design.
Relay Circuits
Relay circuits use a small control current to switch a larger load current, acting as an electrically operated switch. Relays are crucial in automatic control and circuit protection applications. Understanding how relays work and their operational components helps in troubleshooting and designing efficient systems.
In the exercise, a relay with a specific minimum operating current is used, and it was essential to calculate the necessary shunt resistance to maintain proper relay operation given a specific line current.
  • Relays use a small current to control a larger current switch.
  • Ensure minimum operational current for proper relay activation.
  • Shunt resistors help adjust current flow to meet relay requirements.
This particular problem highlights the practical application of relays in managing circuit dynamics, allowing for control over larger current with precision.

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Most popular questions from this chapter

Two resistors, of \(4.00 \Omega\) and \(12.0 \Omega\), are connected in parallel across a 22-V battery having internal resistance \(1.00 \Omega\). Compute (a) the battery current, \((b)\) the current in the \(4.00-\Omega\) resistor, \((c)\) the terminal voltage of the battery, \((d)\) the current in the \(12.0-\Omega\) resistor.

The currents in the circuit are steady. Find \(I_{1}, I_{2}, I_{3}\), and the charge on the capacitor. When a capacitor has a constant charge, as it does here, the current flowing to it is zero. Therefore, \(I_{2}=0\), and the circuit behaves just as though the center wire were missing. With the center wire missing, the remaining circuit is simply 12 \(\Omega\) connected across a 15 -V battery. Therefore, $$ I_{1}=\frac{\varepsilon}{R}=\frac{15 \mathrm{~V}}{12 \Omega}=1.25 \mathrm{~A} $$ In addition, because \(I_{2}=0\), we have \(I_{3}=I_{1}=1.3 \mathrm{~A}\). To find the charge on the capacitor, first find the voltage difference between points- \(a\) and \(-b\). Start at \(a\) and go around the upper path. Voltage change from \(a\) to \(b=-(5.0 \Omega) I_{3}+6.0 \mathrm{~V}+(3.0 \Omega) I_{2}\) $$ =-(5.0 \Omega)(1.25 \mathrm{~A})+6.0 \mathrm{~V}+(3.0 \Omega)(0)=-0.25 \mathrm{~V} $$ Therefore, \(b\) is at the lower potential and the capacitor plate at \(b\) is negative. To find the charge on the capacitor, $$ Q=C V_{a b}=\left(2 \times 10^{-6} \mathrm{~F}\right)(0.25 \mathrm{~V})=0.5 \mu \mathrm{C} $$

What shunt resistance should be connected in parallel with an ammeter having a resistance of \(0.040 \Omega\) so that 25 percent of the total current will pass through the ammeter?

A \(36-\Omega\) galvanometer is shunted by a resistor of \(4.0 \Omega\). What part of the total current will pass through the instrument?

How many \(160-\Omega\) resistors (in parallel) are required to carry a total of \(5.0 \mathrm{~A}\) on a \(100-\mathrm{V}\) line?

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