For the circuit shown in the figure below, the switch S is initially open and the voltage across the capacitor is 60 V. The switch is then closed at time t = 0 s. What is the charge on the capacitor when the current in the circuit is 20 µA? 22 µF 1.5 MQ Ο 140 μC Ο 290 μC Ο 310 μ Ο 970 μC Ο 660 μ ww
For the circuit shown in the figure below, the switch S is initially open and the voltage across the capacitor is 60 V. The switch is then closed at time t = 0 s. What is the charge on the capacitor when the current in the circuit is 20 µA? 22 µF 1.5 MQ Ο 140 μC Ο 290 μC Ο 310 μ Ο 970 μC Ο 660 μ ww
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![## Useful Constants
- Coulomb's constant (\( k \)):
\[
k = 9.00 \times 10^9 \, \frac{N \cdot m^2}{C^2}
\]
- Permittivity of free space (\( \varepsilon_0 \)):
\[
\varepsilon_0 = 8.85 \times 10^{-12} \, \frac{C^2}{N \cdot m^2}
\]
- Elementary charge (\( e \)):
\[
e = 1.6 \times 10^{-19} \, C
\]
- Electron mass (\( m_e \)):
\[
m_e = 9.11 \times 10^{-31} \, kg
\]
- Proton mass (\( m_p \)):
\[
m_p = 1.67 \times 10^{-27} \, kg
\]
- Neutron mass (\( m_n \)):
\[
m_n = 1.68 \times 10^{-27} \, kg
\]
These constants are fundamental in physics, particularly in the fields of electromagnetism and particle physics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a908816-dcaf-4aa4-8dd6-7dd2f0aea84f%2Fd8f15cf1-8b96-40cf-9230-16c20a354135%2Fb5z0qj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Useful Constants
- Coulomb's constant (\( k \)):
\[
k = 9.00 \times 10^9 \, \frac{N \cdot m^2}{C^2}
\]
- Permittivity of free space (\( \varepsilon_0 \)):
\[
\varepsilon_0 = 8.85 \times 10^{-12} \, \frac{C^2}{N \cdot m^2}
\]
- Elementary charge (\( e \)):
\[
e = 1.6 \times 10^{-19} \, C
\]
- Electron mass (\( m_e \)):
\[
m_e = 9.11 \times 10^{-31} \, kg
\]
- Proton mass (\( m_p \)):
\[
m_p = 1.67 \times 10^{-27} \, kg
\]
- Neutron mass (\( m_n \)):
\[
m_n = 1.68 \times 10^{-27} \, kg
\]
These constants are fundamental in physics, particularly in the fields of electromagnetism and particle physics.
![**Circuit Analysis Problem**
For the circuit shown in the diagram, the switch \( S \) is initially open and the voltage across the capacitor is \( 60 \, \text{V} \). The switch is then closed at time \( t = 0 \, \text{s} \). What is the charge on the capacitor when the current in the circuit is \( 20 \, \mu \text{A} \)?
**Circuit Description:**
- Capacitor: \( 22 \, \mu \text{F} \)
- Resistor: \( 1.5 \, \text{M}\Omega \)
**Multiple Choice Answers:**
- \( 140 \, \mu \text{C} \)
- \( 290 \, \mu \text{C} \)
- \( 310 \, \mu \text{C} \)
- \( 970 \, \mu \text{C} \)
- \( 660 \, \mu \text{C} \)
**Solution Explanation:**
To solve for the charge on the capacitor when the current is \( 20 \, \mu \text{A} \), we use the formula for the charging process of an RC circuit:
1. Implement the time constant \(\tau = R \times C\)
2. Use the relationship for charge \( Q \) at a given time with current \( I \):
\[
Q = C \times V
\]
And since the current \( I = \frac{dQ}{dt} \), integrate as needed.
**Note:** This theoretical approach involves understanding the exponential nature of capacitor charging and the application of Ohm's Law (\( V = I \times R \)) in transient analysis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a908816-dcaf-4aa4-8dd6-7dd2f0aea84f%2Fd8f15cf1-8b96-40cf-9230-16c20a354135%2F99tk85k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Circuit Analysis Problem**
For the circuit shown in the diagram, the switch \( S \) is initially open and the voltage across the capacitor is \( 60 \, \text{V} \). The switch is then closed at time \( t = 0 \, \text{s} \). What is the charge on the capacitor when the current in the circuit is \( 20 \, \mu \text{A} \)?
**Circuit Description:**
- Capacitor: \( 22 \, \mu \text{F} \)
- Resistor: \( 1.5 \, \text{M}\Omega \)
**Multiple Choice Answers:**
- \( 140 \, \mu \text{C} \)
- \( 290 \, \mu \text{C} \)
- \( 310 \, \mu \text{C} \)
- \( 970 \, \mu \text{C} \)
- \( 660 \, \mu \text{C} \)
**Solution Explanation:**
To solve for the charge on the capacitor when the current is \( 20 \, \mu \text{A} \), we use the formula for the charging process of an RC circuit:
1. Implement the time constant \(\tau = R \times C\)
2. Use the relationship for charge \( Q \) at a given time with current \( I \):
\[
Q = C \times V
\]
And since the current \( I = \frac{dQ}{dt} \), integrate as needed.
**Note:** This theoretical approach involves understanding the exponential nature of capacitor charging and the application of Ohm's Law (\( V = I \times R \)) in transient analysis.
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