The following two voltage divider circuits are provided to convert 9 V into 6 V. Each resistor used is 100 Ω with ¼ W power rating. Which one of these two circuits would you use to obtain 6 V from 9 V and why? Provide calculations to justify your decision.
The following two voltage divider circuits are provided to convert 9 V into 6 V. Each resistor used is 100 Ω with ¼ W power rating. Which one of these two circuits would you use to obtain 6 V from 9 V and why? Provide calculations to justify your decision.
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter7: Parallel Circuits
Section: Chapter Questions
Problem 3PP: Using the rules for parallel circuits and Ohmslaw, solve for the missing values....
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The following two voltage divider circuits are provided to convert 9 V into 6 V. Each resistor used is 100 Ω with ¼ W power rating. Which one of these two circuits would you use to obtain 6 V from 9 V and why? Provide calculations to justify your decision.
![### Understanding Series and Parallel Circuits
An important aspect of electrical and electronics engineering is the understanding of circuit configurations. The illustration provided shows two common types of resistor circuits: series and parallel circuits, both powered by a 9V battery.
#### Left Diagram: Parallel Circuit
In the left diagram, three resistors, each with a resistance of 100 ohms and a power rating of 1/4 watt, are connected in parallel. Here’s a breakdown of this circuit:
- **Power Source:** 9V battery with positive and negative terminals indicated.
- **Resistors:** Three resistors each labeled "100 Ω, 1/4 W" connected in parallel.
In a parallel circuit, the voltage across each resistor remains the same, which is equal to the voltage of the battery (9V in this case). The overall resistance of resistors in parallel is calculated using the formula:
\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Given the resistances:
\[ \frac{1}{R_{total}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} \]
\[ \frac{1}{R_{total}} = \frac{3}{100} \]
\[ R_{total} = \frac{100}{3} \]
\[ R_{total} \approx 33.33 \, \Omega \]
#### Right Diagram: Series Circuit
In the right diagram, the three resistors are connected in series. Here’s a breakdown of this circuit:
- **Power Source:** 9V battery with positive and negative terminals indicated.
- **Resistors:** Three resistors each labeled "100 Ω, 1/4 W" connected in series.
In a series circuit, the total resistance is simply the sum of the individual resistances:
\[ R_{total} = R_1 + R_2 + R_3 \]
Given the resistances:
\[ R_{total} = 100 + 100 + 100 \]
\[ R_{total} = 300 \, \Omega \]
#### Key Points to Remember
- **Parallel Circuit:** The voltage across each resistor is the same as the source voltage. The total resistance is reduced and calculated using the reciprocal formula.
- **Series Circuit:** The current](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff920003c-b80d-4d46-9199-437e1336b936%2F128ef2d3-2699-4bfe-9773-4c37608f4673%2Fpt32k5_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Series and Parallel Circuits
An important aspect of electrical and electronics engineering is the understanding of circuit configurations. The illustration provided shows two common types of resistor circuits: series and parallel circuits, both powered by a 9V battery.
#### Left Diagram: Parallel Circuit
In the left diagram, three resistors, each with a resistance of 100 ohms and a power rating of 1/4 watt, are connected in parallel. Here’s a breakdown of this circuit:
- **Power Source:** 9V battery with positive and negative terminals indicated.
- **Resistors:** Three resistors each labeled "100 Ω, 1/4 W" connected in parallel.
In a parallel circuit, the voltage across each resistor remains the same, which is equal to the voltage of the battery (9V in this case). The overall resistance of resistors in parallel is calculated using the formula:
\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Given the resistances:
\[ \frac{1}{R_{total}} = \frac{1}{100} + \frac{1}{100} + \frac{1}{100} \]
\[ \frac{1}{R_{total}} = \frac{3}{100} \]
\[ R_{total} = \frac{100}{3} \]
\[ R_{total} \approx 33.33 \, \Omega \]
#### Right Diagram: Series Circuit
In the right diagram, the three resistors are connected in series. Here’s a breakdown of this circuit:
- **Power Source:** 9V battery with positive and negative terminals indicated.
- **Resistors:** Three resistors each labeled "100 Ω, 1/4 W" connected in series.
In a series circuit, the total resistance is simply the sum of the individual resistances:
\[ R_{total} = R_1 + R_2 + R_3 \]
Given the resistances:
\[ R_{total} = 100 + 100 + 100 \]
\[ R_{total} = 300 \, \Omega \]
#### Key Points to Remember
- **Parallel Circuit:** The voltage across each resistor is the same as the source voltage. The total resistance is reduced and calculated using the reciprocal formula.
- **Series Circuit:** The current
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