The schematic is step 7 **Title: Understanding an RC Circuit with Switches****Introduction**In this educational lesson, we will explore a basic RC (Resistor-Capacitor) circuit. The circuit diagram provided demonstrates the integration of a power supply, capacitor, resistor, and two switches, enabling students to understand the charge and discharge cycles of the capacitor within the circuit.**Components in the Circuit**1. **Power Supply (PSB)**: A 5V DC power supply.2. **Capacitor (C1)**: A 100 µF capacitor.3. **Resistor (R1)**: A 10 kΩ resistor.4. **Switches (SW1 and SW2)**: Two switches, SW1 and SW2, are included for controlling the circuit pathways.**Description of the Circuit Diagram**- **PSB 5 V**: - This is the power supply providing a constant voltage of 5V.- **SW1 and SW2**: - There are two switches in the circuit, denoted as SW1 and SW2.- **C1 (100 µF)**: - This represents a capacitor with a capacitance of 100 microfarads.- **R1 (10 kΩ)**: - This represents a resistor with a resistance of 10 kilo-ohms.**Circuit Operation**- **Charging Cycle (with SW1 closed)**: - When switch SW1 is closed and SW2 is open, the circuit allows current to flow from the power supply through the capacitor (C1). The capacitor begins to charge to the voltage provided by the power supply (5V). The charging current is controlled by the resistor (R1), creating a time constant (τ) given by τ = R1 * C1.- **Discharging Cycle (with SW2 closed)**: - When switch SW2 is closed and SW1 is open, the path allows the capacitor to discharge through the resistor R1. The discharge will follow an exponential decay until the voltage across the capacitor reaches zero or very close to zero, depending on the time constant (τ = R1 * C1).**Time Constant (τ)**- The time constant (τ) is the product of the resistance (R) and capacitance (C) in the circuit: τ = R1 * C1 τ = 10 kΩ * 100 µF τ = 10 ### Understanding the Consequences of Closing SW1 and SW2 TogetherIn electrical engineering, certain configurations can lead to various outcomes on your circuit board. One such scenario is when both SW1 and SW2 are closed at the same time. Understanding the consequences is crucial for troubleshooting and safe circuit design.Here are the possible outcomes:1. **Short Circuit Across the PSB**: - **Explanation**: This would create a short circuit across the power supply bus (PSB). This can cause the power LED on the PSB to go out and potentially damage the voltage regulator if the connection remains over an extended period. - **Consideration**: This scenario underscores the importance of ensuring that circuit connections are made properly to avoid damaging components.2. **Rapid Charge and Discharge of the Capacitor**: - **Explanation**: The capacitor will charge and discharge quickly and repeatedly without any LEDs in series on either the charge path or the discharge path. - **Observation**: This behavior is typically not desirable as it can lead to erratic behavior and potential overheating.3. **Charge Neutralization**: - **Explanation**: The paths with equal and opposite charges will cancel each other out, similar to the neutralization observed in a Leyden Jar, an early type of capacitor. - **Safety Note**: An important precaution is to avoid receiving an electric shock during this neutralization process, as it can be unpleasant or dangerous.4. **Simultaneous Charge and Discharge**: - **Explanation**: The capacitor might both charge and discharge simultaneously. Evidence of this process can be observed by measuring the current through both the charging path and the discharging path. - **Insight**: This scenario highlights the complex behavior of circuits where components can have multiple states of operation concurrently.Understanding these outcomes helps in designing safer and more reliable electronic circuits. Proper handling of switches and capacitors ensures efficient operation and prevents damage to components.

Answered: Image /qna-images/answer/5d00f75b-b853-4756-85eb-c66dcfa7f69c.jpg

Answered: The schematic is step 7

The schematic is step 7

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

Similar questions

3- Find IN and RN For the cct. shown at 12 resistar. ? Ansi- IN=0.643A RN = 1-552 1√2 MM 3V 3V2 4√2 202 9₂ 2A
What will be the voltage gain for the given circuit?
1) For each of the three full-adders in Figure 6–6, determine the outputs for the inputs shown. A Σ 國 B Cout Ca 0 A Σ 你 B Cout Cin (a) FIGURE 6-6 0 (b) A B Ca Σ Cout 1 (c)
What are the expected voltages at points C and D of this circuit? 10 0 10 0 12 V 12 V O 12 V and 12 V 2V and 0 V 10 V and 2 V 6 V and 0 V
Assume that RL = 1K Ohm in the loaded voltage divider shown below. Calculate the output voltage across the load resistor from point B to point C using the loaded voltage divider equation. A R1 2k Ohm .9 V R2 RL 1k Ohm 1.8 V 2 V 8V 7.2 V
Root locus None of the choices O is the path traced by poles of a system as K varies from infinity to 0 O branches are always equal to total number of poles and zeros O always starts from zeros and ends at poles
WRITE LEGIBLY AND COMPLETE WORK WITH DRAWINGS. LABEL PROPERLY. NO SHROTCUT. SKIP IF YOU ALREADY DID THIS OR ELSE DOWNVOTE
An R-L series circuit contains two resistors and two inductors. The resistors are 86 kΩ and 68 kΩ. the inductors have and inductive reactances of 24 kΩ and 56 kΩ. The total voltage is 480V. What is the voltage drop across the 56 kΩ inductor?
Given the circuit in Figure 16-5, what is the total circuit current? XL = 300 Ω 600 Vrms O 2.0 MA 1.0 A Vs O 2.24 A O 2.0 A Figure 165 R = 600 Ω
Find equivalent resistance across A and B