/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 How many volts are supplied to o... [FREE SOLUTION] | 91Ó°ÊÓ

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How many volts are supplied to operate an indicator light on a DVD player that has a resistance of \(140 \Omega\) given that \(25.0 \mathrm{mA}\) passes through it?

Short Answer

Expert verified
The voltage supplied to operate the indicator light on the DVD player is \(3.5\,\text{V}\).

Step by step solution

01

Convert current to Amperes

We are given the current in milliamperes (mA). First, we need to convert it to Amperes (A) by dividing the value by 1000. $$ I = 25.0\,\text{mA} * \frac{1\,\text{A}}{1000\,\text{mA}} = 0.025\,\text{A} $$
02

Use Ohm's Law

Now that we have the current in Amperes, we can use Ohm's Law to find the voltage supplied to the indicator light. Ohm's Law is defined as: $$ V = I * R $$ where V is voltage, I is current, and R is resistance.
03

Calculate Voltage

Put the given values of current (0.025 A) and resistance (140 Ω) in the equation: $$ V = 0.025\,\text{A} * 140\,\Omega = 3.5\,\text{V} $$
04

Write the final answer

The voltage supplied to operate the indicator light on the DVD player is 3.5 volts.

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Key Concepts

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

Understanding Electrical Resistance
At its core, electrical resistance can be thought of as a roadblock for electrons moving through a conductor. The higher the resistance, the harder it is for the electrical current to pass through the material. This concept is crucial when working with electronic devices, such as DVD players, as resistance affects how much current will flow at a given voltage.

Measured in ohms \( \Omega \), resistance can be influenced by several factors, including the material's nature, its cross-sectional area, length, and temperature. In the context of the provided exercise, the resistance of the indicator light is given as 140 ohms, which we use to determine how much voltage is needed for the proper operation of the light.

Returns the resistance of a component, it's important to consider that it serves as an intrinsic property dictating how much current the component will draw under a certain voltage. When designing or repairing circuits, understanding resistance ensures components are used within their operational limits to prevent damage.
The Basics of Voltage Calculation
Voltage is the driving force behind the flow of current in a circuit, analogous to the pressure of water in a pipe. Calculating voltage is essential in understanding how electronic devices function and in troubleshooting them when they malfunction.

By using Ohm's Law, we can calculate voltage by multiplying current (I) by resistance (R). In symbol form, the law is expressed as \( V = I \times R \). In the exercise, we're given the resistance (140 \( \Omega \) and the current in milliamperes, which must be converted into amperes as part of the solution process.

Voltage Calculation Example

When a current of 0.025 A—after the conversion from milliamperes—flows through a resistance of 140 \( \Omega \), the voltage is found by substituting these values into Ohm's Law, resulting in a voltage of 3.5 V. It's crucial to input the variables in the correct units (amperes for current, ohms for resistance) to ensure the accuracy of your voltage calculations.
Current Conversion in Electrical Circuits
Electrical current, the flow of electric charge, is typically measured in amperes (A). However, it's often presented in milliamperes (mA) in various applications, necessitating a conversion for practical calculations such as those used in applying Ohm’s Law.

To convert from milliamperes to amperes, divide the current value by 1,000 (since 1 A = 1,000 mA). In our exercise, we start with a current of 25.0 mA. The conversion to amperes is straightforward: \( 25.0\,\text{mA} \times \frac{1\,\text{A}}{1000\,\text{mA}} = 0.025\,\text{A} \).

This conversion is a pivotal step in solving many electrical problems as many components, like the indicator light in our exercise, operate on low-current circuits, and readings and specifications often employ milliamperes. Understanding how to perform this conversion accurately ensures correct application of formulas and avoids potential errors in calculations for voltage and resistance.

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Most popular questions from this chapter

An alternative to CFL. bulbs and incandescent bulbs \(\begin{array}{ll}\text { A } & 100-\mathrm{W}\end{array}\) are light-emitting diode (LED) bulbs. incandescent bulb can be replaced by a 16 -W LED bulb. Both produce 1600 lumens of light. Assuming the cost of electricity is \(\$ 0.10\) per kilowatt-hour, how much does it cost to run the bulb for one year if it runs for four hours a day?

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