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(II) Suppose that you have a \(9.0-\mathrm{V}\) battery and you wish to apply a voltage of only 4.0 \(\mathrm{V}\) . Given an unlimited supply of \(1.0-\Omega\) resistors, how could you connect them so as to make a "voltage divider" that produces a \(4.0-\mathrm{V}\) output for a \(9.0-\mathrm{V}\) input?

Short Answer

Expert verified
Use 5 resistors in series for \( R_1 = 5 \Omega \) and 4 in series for \( R_2 = 4 \Omega \) to create the voltage divider.

Step by step solution

01

Understanding Voltage Divider Concept

A voltage divider is a simple circuit that turns a large voltage into a smaller one. It consists of two resistors in series. The output voltage is taken across the second resistor and is a fraction of the total input voltage.
02

Setting up the Voltage Divider Formula

To find the output voltage (V_{ ext{out}}), we'll use the formula: \( V_{ ext{out}} = V_{ ext{in}} \times \frac{R_2}{R_1 + R_2} \), where \( V_{ ext{in}} \) is the input voltage, and \( R_1 \) and \( R_2 \) are the resistances in the circuit.
03

Input Values into the Formula

With \( V_{ ext{in}} = 9.0 \, \text{V} \) and \( V_{ ext{out}} = 4.0 \, \text{V} \), substitute these into the formula: \( 4.0 = 9.0 \times \frac{R_2}{R_1 + R_2} \).
04

Solving for Resistance Values

Re-arrange the equation: \( 4.0 = 9.0 \times \frac{R_2}{R_1 + R_2} \). This simplifies to \( \frac{R_2}{R_1 + R_2} = \frac{4}{9} \). Solving this gives \( 5R_2 = 4R_1 \).
05

Expressing Resistances as Whole Numbers

To maintain integer numbers, we can choose multiples such that \( R_2 = 4 \times n \) and \( R_1 = 5 \times n \) for some integer \( n \). The simplest choice is \( n = 1 \), giving \( R_1 = 5 \Omega \) and \( R_2 = 4 \Omega \).
06

Connecting Resistors

Using the unlimited supply of \( 1.0-\Omega \) resistors, connect 5 of them in series to form \( R_1 = 5 \Omega \) and 4 of them in series to form \( R_2 = 4 \Omega \). Now, you have a voltage divider that gives a 4.0-V output for a 9.0-V input.

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Key Concepts

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

Resistors in Series
When you have resistors in series, they are connected one after the other in a straight line. This means that the current flows through each resistor one by one. The total or equivalent resistance of resistors in series is just the sum of the individual resistances. So if you have three resistors with resistances of \(1 \text{ }\Omega\), \(2 \text{ }\Omega\), and \(3 \text{ }\Omega\), the total resistance is \(1 + 2 + 3 = 6 \text{ }\Omega\).

In the context of a voltage divider, where resistors are used to break down a voltage into smaller parts, having resistors in series is very handy. You can specifically select how much voltage you want to "drop" across the resistors, simply by changing their values. In our case, using a combination of \(1 \text{ }\Omega\) resistors, we created two larger resistors: \(R_1\) and \(R_2\). This allows us to get the exact output voltage needed.
Ohm's Law
Ohm's Law is one of the most fundamental concepts when dealing with electric circuits. It states that the current \(I\) through a conductor between two points is directly proportional to the voltage \(V\) across the two points and inversely proportional to the resistance \(R\). The formula is given as \(V = IR\).

With this law, you can calculate any one of the three variables (voltage, current, resistance) if the other two are known. In terms of our voltage divider setup, Ohm's Law helps us understand why changing the resistances affects the output voltage.
  • Increased resistance leads to decreased current for the same applied voltage.
  • More current flows through lower resistances when the total voltage remains constant.
It is through manipulating the values of series resistors in a circuit, taking advantage of Ohm's Law, that we crafted our desired voltage difference.
Electric Circuits
Electric circuits involve the flow of electric current through a closed path. They are usually composed of a power source (like a battery), some components (like resistors, capacitors), and interconnections (wires). One of the core applications of electric circuits is controlling the output voltage to match specific requirements, which is what a voltage divider achieves.

In our specific exercise, the electric circuit was composed of a 9V battery and simplified using a series of \(1 \text{ }\Omega\) resistors to create the necessary voltage drop. This was to produce an output voltage of 4V from an input of 9V.
  • The power source provides the total voltage.
  • Resistors adjust output voltage to the desired level.
  • Wire connections help current flow within the circuit.
Managing how this circuit operates allows us to harness and safely distribute electricity, showcasing the strength and flexibility of using electric circuits in practical applications.

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Most popular questions from this chapter

(1) Four \(1.50-\mathrm{V}\) cells are connected in series to a \(12-\Omega\) light bulb. If the resulting current is 0.45 \(\mathrm{A}\) , what is the internal resistance of each cell, assuming they are identical and neglecting the resistance of the wires?

How many \(\frac{1}{2}\) -W resistors, each of the same resistance, must be used to produce an equivalent \(3.2-\mathrm{k} \Omega, 3.5-\mathrm{W}\) resistor? What is the resistance of each, and how must they be connected? Do not exceed \(P=\frac{1}{2} \mathrm{~W}\) in each resistor.

(II) A 45-V battery of negligible internal resistance is connected to a \(44-\mathrm{k} \Omega\) and a \(27-\mathrm{k} \Omega\) resistor in series. What reading will a voltmeter, of internal resistance \(95 \mathrm{k} \Omega,\) give when used to measure the voltage across each resistor? What is the percent inaccuracy due to meter resistance for each case?

(II) A milliammeter reads 25 \(\mathrm{mA}\) full scale. It consists of a \(0.20-\Omega\) resistor in parallel with a \(33-\Omega\) galvanometer. How can you change this ammeter to a voltmeter giving a full scale reading of 25 \(\mathrm{V}\) without taking the ammeter apart? What will be the sensitivity \((\Omega / \mathrm{V})\) of your voltmeter?

(II) An ammeter whose internal resistance is 53\(\Omega\) reads 5.25 \(\mathrm{mA}\) when connected in a circuit containing a battery and two resistors in series whose values are 650\(\Omega\) and 480\(\Omega .\) What is the actual current when the ammeter is absent?

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