The voltage at the terminals of the 0.6 µF capacitor shown in the figure is 0 for t < 0 and 40e-15,000t sin 30, 000t V for t≥ 0. Find (a) i(0); (b) the power delivered to the capacitor at t = 7/80 ms; and (c) the energy stored in the capacitor at t = 7/80 ms.
The voltage at the terminals of the 0.6 µF capacitor shown in the figure is 0 for t < 0 and 40e-15,000t sin 30, 000t V for t≥ 0. Find (a) i(0); (b) the power delivered to the capacitor at t = 7/80 ms; and (c) the energy stored in the capacitor at t = 7/80 ms.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Question
![### Capacitor Voltage and Analysis Problem
The voltage at the terminals of the \(0.6 \, \mu \text{F}\) capacitor shown in the figure is:
\[ v(t) = \begin{cases}
0 & \text{for } t < 0 \\
40 e^{-15,000t} \sin(30,000t) \, \text{V} & \text{for } t \geq 0
\end{cases} \]
You are required to determine the following:
1. \(i(0)\) - the initial current through the capacitor at \( t = 0 \).
2. The power delivered to the capacitor at \( t = \frac{\pi}{80} \, \text{ms} \).
3. The energy stored in the capacitor at \( t = \frac{\pi}{80} \, \text{ms} \).
### Given Data:
- **Capacitance:** \(0.6 \, \mu \text{F}\)
- **Voltage for \( t < 0 \):** \(0 \, \text{V}\)
- **Voltage for \( t \geq 0 \):** \( 40 e^{-15,000t} \sin(30,000t) \, \text{V} \)
### Diagram
The circuit diagram includes:
- A capacitor denoted by a standard capacitor symbol.
- The capacitor is annotated with its capacitance value of \(0.6 \, \mu \text{F}\).
- Voltage across the capacitor \(v\) is shown as \( v \).
- Current through the capacitor \(i\) is shown as \( i \).
- Polarity of the voltage and current direction are indicated by '+' and '-' signs and an arrow respectively.
### Steps to Solve:
**a) Find \( i(0) \):**
Use the relationship for current through a capacitor:
\[
i(t) = C \frac{dv(t)}{dt}
\]
where \( C \) is the capacitance.
**b) Find the Power Delivered to the Capacitor:**
Power \(P(t)\) is given by the product of voltage and current:
\[
P(t) = v(t) \cdot i(t)
\]
Calculate this at \( t = \frac{\pi}{80](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2b6cedf-1de1-433d-8173-2409fffc8f05%2F8d58eb2d-d6ba-46f3-bede-08800f16542a%2Fdfcucl7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Capacitor Voltage and Analysis Problem
The voltage at the terminals of the \(0.6 \, \mu \text{F}\) capacitor shown in the figure is:
\[ v(t) = \begin{cases}
0 & \text{for } t < 0 \\
40 e^{-15,000t} \sin(30,000t) \, \text{V} & \text{for } t \geq 0
\end{cases} \]
You are required to determine the following:
1. \(i(0)\) - the initial current through the capacitor at \( t = 0 \).
2. The power delivered to the capacitor at \( t = \frac{\pi}{80} \, \text{ms} \).
3. The energy stored in the capacitor at \( t = \frac{\pi}{80} \, \text{ms} \).
### Given Data:
- **Capacitance:** \(0.6 \, \mu \text{F}\)
- **Voltage for \( t < 0 \):** \(0 \, \text{V}\)
- **Voltage for \( t \geq 0 \):** \( 40 e^{-15,000t} \sin(30,000t) \, \text{V} \)
### Diagram
The circuit diagram includes:
- A capacitor denoted by a standard capacitor symbol.
- The capacitor is annotated with its capacitance value of \(0.6 \, \mu \text{F}\).
- Voltage across the capacitor \(v\) is shown as \( v \).
- Current through the capacitor \(i\) is shown as \( i \).
- Polarity of the voltage and current direction are indicated by '+' and '-' signs and an arrow respectively.
### Steps to Solve:
**a) Find \( i(0) \):**
Use the relationship for current through a capacitor:
\[
i(t) = C \frac{dv(t)}{dt}
\]
where \( C \) is the capacitance.
**b) Find the Power Delivered to the Capacitor:**
Power \(P(t)\) is given by the product of voltage and current:
\[
P(t) = v(t) \cdot i(t)
\]
Calculate this at \( t = \frac{\pi}{80
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Answer to B and C is incorrect. Please fix.
Could you show me the correct steps and how to get the correct answer.
The correct answers are:
B: -649.2mW
C: 126.13 uJ
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