The voltage across a 21-uF capacitor is given by the following function of time, vo Find the Power in Watts at time t-9.5. (The angle is in radians) v(t) = 200 sin(500t) i(t) v(t)
The voltage across a 21-uF capacitor is given by the following function of time, vo Find the Power in Watts at time t-9.5. (The angle is in radians) v(t) = 200 sin(500t) i(t) v(t)
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![## Capacitors in AC Circuits: Calculating Power
### Problem Statement:
The voltage across a 21-µF capacitor is given by the following function of time, \( v(t) = 200 \sin(500t) \) V.
Find the Power in Watts at time \( t = 9.5 \). (The angle is in radians)
### Given Voltage Function:
\[ v(t) = 200 \sin(500t) \]
### Capacitor Circuit Diagram:
The diagram shows:
- A capacitor, denoted by \( C \).
- The voltage across the capacitor \( v(t) \) with positive and negative terminals.
- The current through the capacitor \( i(t) \).
### Solution Steps:
To find the power, we need to calculate the instantaneous power using the formula:
\[ P(t) = v(t) \cdot i(t) \]
1. **Determine the current \( i(t) \) through the capacitor.**
For a capacitor, the current \( i(t) \) is related to the voltage \( v(t) \) by:
\[ i(t) = C \frac{d}{dt} v(t) \]
2. **Differentiate the voltage function to find \( i(t) \):**
Given \( v(t) = 200 \sin(500t) \):
\[ \frac{d}{dt} v(t) = 200 \cdot 500 \cos(500t) \]
\[ i(t) = 21 \times 10^{-6} \cdot 100,000 \cos(500t) \]
\[ i(t) = 2.1 \cos(500t) \]
3. **Substitute \( t = 9.5 \) into the expressions for \( v(t) \) and \( i(t) \):**
\[ v(9.5) = 200 \sin(500 \cdot 9.5) \]
\[ i(9.5) = 2.1 \cos(500 \cdot 9.5) \]
4. **Calculate \( P(9.5) \):**
\[ P(9.5) = v(9.5) \cdot i(9.5) \]
### Final Calculation:
With the specific numeric time \( t = 9.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafc8379c-b6a2-4db8-9fce-fab4471db5da%2F89323607-1a8c-45e0-bb31-94fe733a2f7f%2Fcm3ya7r_processed.png&w=3840&q=75)
Transcribed Image Text:## Capacitors in AC Circuits: Calculating Power
### Problem Statement:
The voltage across a 21-µF capacitor is given by the following function of time, \( v(t) = 200 \sin(500t) \) V.
Find the Power in Watts at time \( t = 9.5 \). (The angle is in radians)
### Given Voltage Function:
\[ v(t) = 200 \sin(500t) \]
### Capacitor Circuit Diagram:
The diagram shows:
- A capacitor, denoted by \( C \).
- The voltage across the capacitor \( v(t) \) with positive and negative terminals.
- The current through the capacitor \( i(t) \).
### Solution Steps:
To find the power, we need to calculate the instantaneous power using the formula:
\[ P(t) = v(t) \cdot i(t) \]
1. **Determine the current \( i(t) \) through the capacitor.**
For a capacitor, the current \( i(t) \) is related to the voltage \( v(t) \) by:
\[ i(t) = C \frac{d}{dt} v(t) \]
2. **Differentiate the voltage function to find \( i(t) \):**
Given \( v(t) = 200 \sin(500t) \):
\[ \frac{d}{dt} v(t) = 200 \cdot 500 \cos(500t) \]
\[ i(t) = 21 \times 10^{-6} \cdot 100,000 \cos(500t) \]
\[ i(t) = 2.1 \cos(500t) \]
3. **Substitute \( t = 9.5 \) into the expressions for \( v(t) \) and \( i(t) \):**
\[ v(9.5) = 200 \sin(500 \cdot 9.5) \]
\[ i(9.5) = 2.1 \cos(500 \cdot 9.5) \]
4. **Calculate \( P(9.5) \):**
\[ P(9.5) = v(9.5) \cdot i(9.5) \]
### Final Calculation:
With the specific numeric time \( t = 9.
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