An R=131k resistor and a C=26μF capacitor are connected in series with a open switch. The capacitor is initially fully charged to an initial potential of V₁ = 21V. The switch is closed at t = 0. Determine the current in the circuit when the charge on the capacitor is 1/7 of its maximum value (in μA).
An R=131k resistor and a C=26μF capacitor are connected in series with a open switch. The capacitor is initially fully charged to an initial potential of V₁ = 21V. The switch is closed at t = 0. Determine the current in the circuit when the charge on the capacitor is 1/7 of its maximum value (in μA).
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
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An R=131k52 resistor and a C =26. capacitor are connected in series with a open switch. The capacitor is initially fully charged to an initial
potential of Vo =21V. The switch is closed att = 0. Determine the
current in the circuit when
the charge on the capacitor is 1/7 of its maximum value
Could you please look at correct answer and show all work
![**Transcription and Explanation for Educational Website**
---
**Problem Statement:**
An \( R = 131 \, k\Omega \) resistor and a \( C = 26 \, \mu F \) capacitor are connected in series with an open switch. The capacitor is initially fully charged to an initial potential of \( V_0 = 21 \, V \). The switch is closed at \( t = 0 \). Determine the current in the circuit when the charge on the capacitor is \( \frac{1}{7} \) of its maximum value (in \(\mu A\)).
**Your Answer:**
17.07
**Correct Answer:**
22.9008 (margin of error +/- 1%)
---
**Explanation:**
This problem involves analyzing an RC (resistor-capacitor) circuit where the capacitor is initially charged, and you need to determine the current when the charge drops to a specific fraction of its initial value.
**Key Concepts:**
1. **RC Circuit Dynamics:**
- When the switch is closed, the circuit begins to discharge.
- The current (\( I(t) \)) and the charge (\( Q(t) \)) on the capacitor change over time based on exponential decay.
2. **Capacitor Charge Equation:**
- \( Q(t) = Q_0 \cdot e^{-t/(RC)} \) where \( Q_0 = C \cdot V_0 \).
3. **Current Equation:**
- \( I(t) = \frac{V_0}{R} \cdot e^{-t/(RC)} \).
4. **Given Condition:**
- The problem states \( Q(t) = \frac{1}{7}Q_0 \).
- Use this condition to find \( t \) and subsequently find \( I(t) \).
5. **Steps to Solve:**
- Compute \( Q_0 = C \cdot V_0 \).
- Substitute \( Q(t) = \frac{1}{7}Q_0 \) into \( Q(t) = Q_0 \cdot e^{-t/(RC)} \) to solve for \( t \).
- Substitute \( t \) back into \( I(t) = \frac{V_0}{R} \cdot e^{-t/(RC)} \) to find the current at that time.
This exercise](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee22cf2f-b974-4b00-a3cf-09b388e7d65d%2F08642129-f9df-4d4a-9304-90bee093e168%2Feh9l0hn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription and Explanation for Educational Website**
---
**Problem Statement:**
An \( R = 131 \, k\Omega \) resistor and a \( C = 26 \, \mu F \) capacitor are connected in series with an open switch. The capacitor is initially fully charged to an initial potential of \( V_0 = 21 \, V \). The switch is closed at \( t = 0 \). Determine the current in the circuit when the charge on the capacitor is \( \frac{1}{7} \) of its maximum value (in \(\mu A\)).
**Your Answer:**
17.07
**Correct Answer:**
22.9008 (margin of error +/- 1%)
---
**Explanation:**
This problem involves analyzing an RC (resistor-capacitor) circuit where the capacitor is initially charged, and you need to determine the current when the charge drops to a specific fraction of its initial value.
**Key Concepts:**
1. **RC Circuit Dynamics:**
- When the switch is closed, the circuit begins to discharge.
- The current (\( I(t) \)) and the charge (\( Q(t) \)) on the capacitor change over time based on exponential decay.
2. **Capacitor Charge Equation:**
- \( Q(t) = Q_0 \cdot e^{-t/(RC)} \) where \( Q_0 = C \cdot V_0 \).
3. **Current Equation:**
- \( I(t) = \frac{V_0}{R} \cdot e^{-t/(RC)} \).
4. **Given Condition:**
- The problem states \( Q(t) = \frac{1}{7}Q_0 \).
- Use this condition to find \( t \) and subsequently find \( I(t) \).
5. **Steps to Solve:**
- Compute \( Q_0 = C \cdot V_0 \).
- Substitute \( Q(t) = \frac{1}{7}Q_0 \) into \( Q(t) = Q_0 \cdot e^{-t/(RC)} \) to solve for \( t \).
- Substitute \( t \) back into \( I(t) = \frac{V_0}{R} \cdot e^{-t/(RC)} \) to find the current at that time.
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