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Chapter 3: Problem 13

Prior to \(t=0, \mathrm{a} 100-\mu \mathrm{F}\) capacitance is uncharged. Starting at \(t=0\), the voltage across the capacitor is increased linearly with time to \(100 \mathrm{~V}\) in \(2 \mathrm{~s}\). Then, the voltage remains constant at \(100 \mathrm{~V}\). Sketch the voltage, current, power, and stored energy to scale versus time.

Short Answer

Expert verified
Voltage increases linearly to 100 V in 2 s, current is 5 mA for 2 s then 0. Power increases to 0.5 W then 0, and energy increases to 0.5 J then constant.

Step by step solution

01

Sketch the Voltage vs. Time

The voltage increases linearly from 0 V at \(t=0\) to 100 V at \(t=2\) seconds. After reaching 100 V, it remains constant. Thus, the voltage vs. time graph is a straight line from \((0, 0)\) to \((2, 100)\) followed by a horizontal line until the time ends.
02

Determine the Current vs. Time

The current \(I(t)\) in a capacitor is given by \(I(t) = C \cdot \frac{dV(t)}{dt}\). For \(0 \leq t \leq 2\), \(V(t) = 50t\), so \(I(t) = 100 \times 10^{-6} \cdot 50 = 5 \times 10^{-3}\) A. For \(t > 2\), the derivative of \(V(t)\) is 0, hence the current is 0 A.
03

Sketch the Current vs. Time

The current is 5 mA (from Step 2) from \(t=0\) to \(t=2\). At \(t > 2\), the current drops to 0 A. Thus, the current vs. time graph is a horizontal line at 5 mA from \(t=0\) to \(t=2\) and a horizontal line at 0 A for \(t > 2\).
04

Calculate Power vs. Time

Power \(P(t)\) in a capacitor is given by \(P(t) = V(t) \cdot I(t)\). For \(0 \leq t \leq 2\), since \(V(t) = 50t\) and \(I(t) = 5 \times 10^{-3}\) A, \(P(t) = 50t \times 5 \times 10^{-3} = 0.25t\) W. For \(t > 2\), \(P(t) = 100 \times 0 = 0\) W.
05

Sketch the Power vs. Time

From Step 4, power increases linearly from \(0\) at \(t=0\) to \(0.5 \) W at \(t=2\) s. After \(t=2\), power drops to 0 W and remains there. The power vs. time graph is a sloped line from \(0\) to \(2\) s, then a horizontal line at 0 W.
06

Compute Stored Energy vs. Time

The energy \(E(t)\) stored in the capacitor is \(E(t) = \frac{1}{2}C[V(t)]^2\). For \(0 \leq t \leq 2\), \(E(t) = \frac{1}{2} \times 100 \times 10^{-6} \times (50t)^2 = 0.125t^2\) J. For \(t > 2\), energy remains \(\frac{1}{2} \times 100 \times 10^{-6} \times (100)^2 = 0.5\) J.
07

Sketch the Stored Energy vs. Time

From Step 6, plot the energy as a parabola growing as \(0.125t^2\) from \(t=0\) to \(t=2\), reaching 0.5 J. At \(t > 2\), it stays constant at 0.5 J.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Voltage vs. Time Graph
Understanding a Voltage vs. Time Graph is fundamental when analyzing capacitor behavior, especially during charging. When the voltage across a capacitor is increased linearly, as in this exercise, the graph looks like a straightforward diagonal line starting from the origin. Here, the voltage increases from 0 V at the start (\(t=0\)), reaching 100 V at 2 seconds, and then remains unchanged.

Key points for a linear increase:
  • The graph is a straight line from \(t=0\) to \(t=2\) seconds.
  • After \(t=2\), the graph becomes a horizontal line, indicating a constant voltage.
By visualizing this line on the graph, it becomes clear how the capacitor voltage ramps up steadily until it stabilizes.
Current vs. Time Graph
For a capacitor, the Current vs. Time Graph illustrates how the current changes as the capacitor charges. The relationship between current and voltage in a capacitor is given by \(I(t) = C \cdot \frac{dV(t)}{dt}\). This means that the current is directly related to the rate of change of the voltage.

In this scenario:
  • The current is constant at 5 mA from \(t=0\) to \(t=2\).
  • Once the voltage stops increasing (after \(t=2\)), the current drops to 0 A.
These features result in a graph that has:
  • A horizontal line from \(t=0\) to \(t=2\) at 5 mA.
  • A horizontal line at 0 A for \(t>2\).
This change reflects the fact that when the voltage becomes steady, there is no further current required to charge the capacitor.
Power vs. Time Calculation
Calculating Power vs. Time in a capacitor helps us understand energy consumption during the charging process. The power \(P(t)\) is determined by multiplying voltage \(V(t)\) by current \(I(t)\). During this challenge:

From \(t=0\) to \(t=2\):
  • \(P(t) = V(t) \cdot I(t) = 50t \cdot 5 \times 10^{-3} = 0.25t\) W, showing a steady increase.
  • At \(t=2\), this results in a power of \(0.5\) W.
After \(t=2\):
  • With constant voltage, power becomes zero \(P(t) = 100 \times 0 = 0\) W.
The power graph starts with a slope to reflect the power increase over time and then levels off once the capacitor stops charging.
Stored Energy in a Capacitor
Exploring Stored Energy in a capacitor allows us to see how energy builds up as the capacitor charges. The formula \(E(t) = \frac{1}{2}C[V(t)]^2\) captures this process.

From \(t=0\) to \(t=2\):
  • Here, the stored energy increases as a parabola since the energy depends on \((50t)^2\), or \(E(t) = 0.125t^2\) J.
At \(t=2\) and beyond:
  • The stored energy reaches a maximum of 0.5 J and stays there.
This suggests that before reaching full charge, the energy accumulation is non-linear, while after stabilization, the energy remains constant as the capacitor holds its charge efficiently.

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Most popular questions from this chapter

What is the equivalent circuit element for a very large (ideally, infinite) inductance having an initial current of \(10 \mathrm{~A}\). Explain your answer.

The maximum value obtained by connecting 4 identical inductances in series is \(16 \mathrm{H}\). What is the value of each inductor? What is the minimum value achieved when they are connected in parallel?

Consider a very large capacitance (ideally, infinite) charged to \(10 \mathrm{~V}\). What other circuit element has the same current-voltage relationship? Explain your answer.

Explain, in a few sentences, the fluid-flow analogy for an inductor.

The current through a 200-mH inductance is given by \(i_{L}(t)=\exp (-2 t) \sin (4 \pi t)\) A in which the angle is in radians. Using your knowledge of calculus, find an expression for the voltage across the inductance. Then, use MATLAB to verify your answer for the voltage and to plot both the current and the voltage for \(0 \leq\) \(t \leq 2 \mathrm{~s}\).
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