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Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
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...
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In the circuit of the figure below, the switch S has been open for a long time. It is then suddenly closed. Take  = 10.0 V, R1 = 43.0 kΩ, R2 = 180 kΩ, and C = 14.0 µF.

(a) Determine the time constant before the switch is closed.

(b) Determine the time constant after the switch is closed.

(c) Let the switch be closed at t = 0. Determine the current in the switch as a function of time. (Assume I is in A and t is in s. Do not enter units in your expression. Use the following as necessary: t.)

I keep messing up part C.  any help you can give and explain would be appreciated. 

 

The image depicts a simple electrical circuit diagram consisting of the following components: 1. **Battery (\( \mathcal{E} \))**: Represented on the left side, it provides the electromotive force (EMF) for the circuit. 2. **Resistors (\( R_1 \) and \( R_2 \))**: These are represented by the zigzag symbols. \( R_1 \) is positioned on the upper branch of the circuit, while \( R_2 \) is on the lower branch. 3. **Switch (S)**: Located in the middle branch, indicated by a diagonal line intersecting a break in the circuit. This switch can open or close the circuit. 4. **Capacitor (C)**: Shown on the right side of the circuit, marked with the parallel lines symbol. The overall circuit forms a closed loop when the switch (S) is closed, enabling current to flow. The configuration suggests that when the switch is closed, the current can pass through \( R_1 \) and \( R_2 \), and charge the capacitor \( C \). This diagram illustrates the fundamental components of a basic RC (resistor-capacitor) circuit used in electrical and electronics engineering for analyzing charge and discharge behaviors in capacitors.
Transcribed Image Text:The image depicts a simple electrical circuit diagram consisting of the following components: 1. **Battery (\( \mathcal{E} \))**: Represented on the left side, it provides the electromotive force (EMF) for the circuit. 2. **Resistors (\( R_1 \) and \( R_2 \))**: These are represented by the zigzag symbols. \( R_1 \) is positioned on the upper branch of the circuit, while \( R_2 \) is on the lower branch. 3. **Switch (S)**: Located in the middle branch, indicated by a diagonal line intersecting a break in the circuit. This switch can open or close the circuit. 4. **Capacitor (C)**: Shown on the right side of the circuit, marked with the parallel lines symbol. The overall circuit forms a closed loop when the switch (S) is closed, enabling current to flow. The configuration suggests that when the switch is closed, the current can pass through \( R_1 \) and \( R_2 \), and charge the capacitor \( C \). This diagram illustrates the fundamental components of a basic RC (resistor-capacitor) circuit used in electrical and electronics engineering for analyzing charge and discharge behaviors in capacitors.
### Analysis of an RC Circuit with a Switch In the circuit diagram, we have a series RC (resistor-capacitor) circuit with a switch \( S \) that has been open for a long time before it is suddenly closed. The following parameters are given: - Electromotive Force (\( \mathcal{E} \)) = 10.0 V - Resistor \( R_1 \) = 43.0 kΩ - Resistor \( R_2 \) = 180 kΩ - Capacitor \( C \) = 14.0 μF #### Questions & Solutions (a) **Determine the time constant before the switch is closed.** - The time constant (\( \tau \)) for an RC circuit is given by the formula: \[ \tau = R \times C \] - When the switch is open, the relevant resistor in the circuit is \( R_1 \). - Thus, the time constant before the switch is closed is: \[ \tau = R_1 \times C = 43.0 \, \text{k}\Omega \times 14.0 \, \mu\text{F} = 3.122 \, \text{s} \] (b) **Determine the time constant after the switch is closed.** - After the switch is closed, the resistors \( R_1 \) and \( R_2 \) are in parallel. The effective resistance \( R \) is given by: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \] - Calculate \( R \) and then find the new time constant: \[ R = \left(\frac{1}{43.0} + \frac{1}{180}\right)^{-1} \, \text{k}\Omega \] \[ \tau = R \times C = 2.52 \, \text{s} \] (c) **Determine the current in the switch as a function of time after the switch is closed.** - The expression for the current \( I(t) \) in the switch as a function of time \( t \) is: \[ I = 252 + 55.5e^{-\frac{t}{2.
Transcribed Image Text:### Analysis of an RC Circuit with a Switch In the circuit diagram, we have a series RC (resistor-capacitor) circuit with a switch \( S \) that has been open for a long time before it is suddenly closed. The following parameters are given: - Electromotive Force (\( \mathcal{E} \)) = 10.0 V - Resistor \( R_1 \) = 43.0 kΩ - Resistor \( R_2 \) = 180 kΩ - Capacitor \( C \) = 14.0 μF #### Questions & Solutions (a) **Determine the time constant before the switch is closed.** - The time constant (\( \tau \)) for an RC circuit is given by the formula: \[ \tau = R \times C \] - When the switch is open, the relevant resistor in the circuit is \( R_1 \). - Thus, the time constant before the switch is closed is: \[ \tau = R_1 \times C = 43.0 \, \text{k}\Omega \times 14.0 \, \mu\text{F} = 3.122 \, \text{s} \] (b) **Determine the time constant after the switch is closed.** - After the switch is closed, the resistors \( R_1 \) and \( R_2 \) are in parallel. The effective resistance \( R \) is given by: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \] - Calculate \( R \) and then find the new time constant: \[ R = \left(\frac{1}{43.0} + \frac{1}{180}\right)^{-1} \, \text{k}\Omega \] \[ \tau = R \times C = 2.52 \, \text{s} \] (c) **Determine the current in the switch as a function of time after the switch is closed.** - The expression for the current \( I(t) \) in the switch as a function of time \( t \) is: \[ I = 252 + 55.5e^{-\frac{t}{2.
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