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).

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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**

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**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%)

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**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
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. This exercise
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