2. Now assume that C'eq = 10 μF and the voltage across this new equivalent capacitance is v (t) = 5 (1-e-4000t) V fort > 0. For this question, answer parts a and b below. a. The energy stored in the new equivalent capacitance at t = 0 ms is w (t) = b. The energy stored in the new equivalent capacitance at t= 10 ms is w (t) = m.J
2. Now assume that C'eq = 10 μF and the voltage across this new equivalent capacitance is v (t) = 5 (1-e-4000t) V fort > 0. For this question, answer parts a and b below. a. The energy stored in the new equivalent capacitance at t = 0 ms is w (t) = b. The energy stored in the new equivalent capacitance at t= 10 ms is w (t) = m.J
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Chapter1: Units, Trigonometry. And Vectors
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The circuit diagram presents a combination of capacitors arranged in a specific configuration. The capacitors are denoted by the symbol "C" and their arrangement forms a more complex equivalent capacitance, \(C_{eq}\).
#### Tasks and Questions:
1. **Equivalent Capacitance Calculation:**
Assume that \(C = 90 \, \text{mF}\). Find the equivalent capacitance across the terminals on the left and fill in the blank with the answer \(C_{eq}\).
\[ C_{eq} = \underline{\hspace{50px}} \, \text{mF} \]
2. **Energy Stored in the Capacitor:**
Now assume that \(C_{eq} = 10 \, \mu\text{F}\) and the voltage across this new equivalent capacitance is \( v(t) = 5 \left( 1 - e^{-4000t} \right) \) V for \( t \geq 0 \). For this question, answer parts a and b below.
a. **Energy stored at \( t = 0 \text{ ms} \):**
\[ w(t) = \underline{\hspace{100px}} \, \text{J} \]
b. **Energy stored at \( t = 10 \text{ ms} \):**
\[ w(t) = \underline{\hspace{100px}} \, \text{mJ} \]
#### Explanation of the Diagram:
- The diagram consists of two sections of capacitors.
- These capacitors are arranged in a combination of series and parallel.
- The goal is to simplify the configuration to find the total or equivalent capacitance (\(C_{eq}\)).
### Step-by-Step Solution:
1. **Analyzing the Top Section of the Circuit:**
- Identify capacitors in series and parallel configurations and apply the respective formulas for equivalent capacitances.
- For capacitors in series:
\[
\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}
\]
- For capacitors in parallel:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07dc3289-c262-41f1-9fc5-a8d18500de1b%2F39071d1f-26be-457a-9b9f-335071e37adf%2Fzhwiw8c_processed.png&w=3840&q=75)
Transcribed Image Text:### Circuit Analysis - Capacitance and Energy Storage
**Consider the circuit shown below:**
The circuit diagram presents a combination of capacitors arranged in a specific configuration. The capacitors are denoted by the symbol "C" and their arrangement forms a more complex equivalent capacitance, \(C_{eq}\).
#### Tasks and Questions:
1. **Equivalent Capacitance Calculation:**
Assume that \(C = 90 \, \text{mF}\). Find the equivalent capacitance across the terminals on the left and fill in the blank with the answer \(C_{eq}\).
\[ C_{eq} = \underline{\hspace{50px}} \, \text{mF} \]
2. **Energy Stored in the Capacitor:**
Now assume that \(C_{eq} = 10 \, \mu\text{F}\) and the voltage across this new equivalent capacitance is \( v(t) = 5 \left( 1 - e^{-4000t} \right) \) V for \( t \geq 0 \). For this question, answer parts a and b below.
a. **Energy stored at \( t = 0 \text{ ms} \):**
\[ w(t) = \underline{\hspace{100px}} \, \text{J} \]
b. **Energy stored at \( t = 10 \text{ ms} \):**
\[ w(t) = \underline{\hspace{100px}} \, \text{mJ} \]
#### Explanation of the Diagram:
- The diagram consists of two sections of capacitors.
- These capacitors are arranged in a combination of series and parallel.
- The goal is to simplify the configuration to find the total or equivalent capacitance (\(C_{eq}\)).
### Step-by-Step Solution:
1. **Analyzing the Top Section of the Circuit:**
- Identify capacitors in series and parallel configurations and apply the respective formulas for equivalent capacitances.
- For capacitors in series:
\[
\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}
\]
- For capacitors in parallel:
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