The Current Through A 0.5-f Capacitor Is 6(1-e-t)a. Determine The Voltage And Power At T=2 S. Assume
Understanding the behavior of capacitors in electrical circuits is fundamental for engineers and students alike. When analyzing a capacitor's response to a given current, it is crucial to understand how voltage and power are related to current over time. This article explores the problem of determining the voltage and power across a 0.5-f capacitor at a specific time, given a prescribed current function. We will walk through the fundamental concepts, detailed calculations, and implications of these results, providing a comprehensive guide for both theoretical understanding and practical applications.
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Introduction to Capacitors and Their Behavior in Circuits
Capacitors are passive electronic components that store electrical energy in an electric field. They are widely used in various applications, including filtering, energy storage, and signal processing. The fundamental relationship governing a capacitor is:
\[ i(t) = C \frac{d v(t)}{dt} \]
where:
- \( i(t) \) is the instantaneous current through the capacitor,
- \( v(t) \) is the voltage across the capacitor,
- \( C \) is the capacitance.
Understanding this relationship allows us to determine the voltage across the capacitor when the current is known, and vice versa.
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Problem Statement and Given Data
The problem provides the current through a 0.5-f capacitor as a function of time:
\[
i(t) = 6(1 - e^{-t})\, \text{A}
\]
where:
- \( C = 0.5\, \text{F} \),
- \( t \) is in seconds.
The task is to:
- Determine the voltage \( v(t) \) across the capacitor at \( t = 2\, \text{seconds} \).
- Calculate the instantaneous power \( p(t) \) at \( t = 2\, \text{seconds} \).
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Fundamental Equations and Approach
To find the voltage and power at a specific time, we need to:
- Integrate the current over time to find the voltage,
- Use the relationship between current and voltage in a capacitor,
- Calculate power as the product of instantaneous voltage and current.
The key equations are:
\[
v(t) = \frac{1}{C} \int_{0}^{t} i(\tau) d\tau + v(0)
\]
Assuming the initial voltage \( v(0) = 0 \) (the capacitor initially uncharged), this simplifies the analysis.
The power at any time is:
\[
p(t) = v(t) \times i(t)
\]
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Calculating the Voltage \( v(t) \) at \( t=2\, \text{seconds} \)
Step 1: Integrate the Current to Find Voltage
Given the initial voltage \( v(0) = 0 \), the voltage across the capacitor is:
\[
v(t) = \frac{1}{C} \int_{0}^{t} i(\tau) d\tau
\]
Substitute \( i(\tau) = 6(1 - e^{-\tau}) \):
\[
v(t) = \frac{1}{0.5} \int_{0}^{t} 6(1 - e^{-\tau}) d\tau
\]
Simplify the constant:
\[
v(t) = 2 \times 6 \int{0}^{t} (1 - e^{-\tau}) d\tau = 12 \int{0}^{t} (1 - e^{-\tau}) d\tau
\]
Step 2: Calculate the Integral
Compute:
\[
\int{0}^{t} (1 - e^{-\tau}) d\tau = \int{0}^{t} 1 d\tau - \int_{0}^{t} e^{-\tau} d\tau
\]
which yields:
\[
t - \left[ -e^{-\tau} \right]_0^t = t - \left( -e^{-t} + e^{0} \right) = t - ( -e^{-t} + 1 ) = t + e^{-t} - 1
\]
Therefore:
\[
v(t) = 12 \left( t + e^{-t} - 1 \right)
\]
Step 3: Find \( v(2) \)
Plug in \( t=2 \):
\[
v(2) = 12 \left( 2 + e^{-2} - 1 \right) = 12 (1 + e^{-2})
\]
Recall that \( e^{-2} \approx 0.1353 \):
\[
v(2) \approx 12 (1 + 0.1353) = 12 \times 1.1353 \approx 13.6236\, \text{V}
\]
Result:
\[
\boxed{
v(2) \approx 13.62\, \text{V}
}
\]
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Calculating the Power \( p(t) \) at \( t=2\, \text{seconds} \)
Step 1: Recall the current at \( t=2 \)
\[
i(2) = 6(1 - e^{-2}) \approx 6 \times (1 - 0.1353) = 6 \times 0.8647 \approx 5.188\, \text{A}
\]
Step 2: Use the voltage and current to find power
Power is:
\[
p(t) = v(t) \times i(t)
\]
At \( t=2 \):
\[
p(2) = 13.62\, \text{V} \times 5.188\, \text{A} \approx 70.66\, \text{W}
\]
Result:
\[
\boxed{
p(2) \approx 70.66\, \text{W}
}
\]
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Discussion of Results
The calculations reveal that at \( t=2\, \text{seconds} \):
- The voltage across the capacitor is approximately 13.62 volts.
- The instantaneous power dissipated or stored is approximately 70.66 watts.
These values reflect the capacitor's charging process, where the current initially is higher and decreases exponentially, while the voltage builds up over time.
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Additional Insights and Practical Considerations
1. Behavior Over Time
- As \( t \to \infty \), \( e^{-t} \to 0 \), and the voltage approaches:
but since the current approaches 6 A asymptotically, the voltage will tend to a steady state. In fact, the integral indicates that the voltage stabilizes at a finite value, considering the initial conditions and the nature of the current.
2. Power Considerations
- The power calculated is instantaneous; in real circuits, power dissipation would also depend on resistive elements.
- Since the analysis assumes an ideal capacitor with no resistance, the power here represents energy stored in the electric field.
3. Application in Circuit Design
- Understanding the voltage and power at specific times helps in designing circuits that require precise timing and energy management.
- For example, in pulse circuits, knowing how quickly a capacitor charges is critical for timing applications.
4. Limitations and Assumptions
- The initial voltage \( v(0) \) was assumed to be zero.
- The calculations assume ideal components, neglecting parasitic resistances and inductances, which would affect real-world behavior.
Conclusion
This detailed analysis demonstrates how to determine the voltage across a capacitor and the instantaneous power at a specific time, given a known current function. By integrating the current over time and applying fundamental capacitor equations, engineers can predict the behavior of capacitive components in various circuits. The approach outlined here—integrating current to find voltage and multiplying voltage by current for power—is fundamental in electrical engineering and essential for designing and analyzing dynamic systems involving capacitors.
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Summary of Key Results
- Voltage at \( t=2\, \text{seconds} \): approximately 13.62 volts
- Power at \( t=2\, \text{seconds} \): approximately 70.66 watts
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Meta Description:
Learn how to determine the voltage and power across a 0.5-f capacitor at a specific time given a current function. This comprehensive guide covers integral calculations, practical implications, and key concepts in circuit analysis.
