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Chapter 24: Problem 34

If a capacitor initially holds no charge, the voltage across it as a function of time can be computed as $$V(t)=\frac{1}{C} \int_{0}^{t} i(t) d t$$ If \(C=10^{-5}\) farad, use the following current data to develop a plot of voltage versus time: $$\begin{array}{c|cccccc} t, s & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 & 1.2 \\ \hline i_{i}, 10^{-3} \mathrm{A} & 0.2 & 0.3683 & 0.3819 & 0.2282 & 0.0486 & 0.0082 & 0.1441 \end{array}$$

Short Answer

Expert verified
The calculated voltages at each time interval are as follows: - For t = 0 seconds: \(V(0) \approx 2 \times 10^{-3}\) V - For t = 0.2 seconds: \(V(0.2) \approx 5.6815 \times 10^{-2}\) V - (Continue calculating voltage for other time intervals) Using these voltage values, plot the voltage versus time graph to observe the increase of voltage as the current passes through the capacitor.

Step by step solution

01

Analyze the given equation

We're given the equation that relates voltage, current, and time for a capacitor: \(V(t)=\frac{1}{C} \int_{0}^{t} i(t) d t\). In this equation, V(t) is the voltage across the capacitor as a function of time, C is the capacitance in farads, and i(t) is the current as a function of time.
02

Identify the capacitance value

We are given that the capacitance of the capacitor, C, is equal to \(10^{-5}\) farad. So, we can plug this value into the equation when calculating the voltage for each time interval.
03

Integrate the current function

Our goal is to find the voltage for each time interval given in the table. Since we have discrete values of current for specific time intervals, we can approximate the integration using the composite trapezoidal rule. The trapezoidal rule for integration states that if we know the current value at each time interval, we can write down the integral as follows: \(\int_{0}^{t} i(t') d t' \approx \Delta t \left( \frac{i_{0}}{2} + i_{1} + i_{2} + \dots + i_{n-1} + \frac{i_{n}}{2} \right)\). The trapezoidal rule approximates the integral by breaking the area under the curve into trapezoids.
04

Calculate the voltage at each time interval

Using the trapezoidal rule for integration and the capacitance value, we can calculate the voltage at each time interval by applying the equation \(V(t) = \frac{1}{C} (\text{integral of current over time})\). We'll calculate the voltage for each time interval given by the table: - For t = 0 seconds: - \(\Delta t = 0.2\); \(i_{0} = 0.2 \times 10^{-3}\) - Voltage: \(V(0) = \frac{1}{10^{-5}} \times 0.2 \times \frac{0.2 \times 10^{-3}}{2} \approx 2 \times 10^{-3}\) V - For t = 0.2 seconds: - \(\Delta t = 0.2\); \(i_{1} = 0.3683 \times 10^{-3}\) - Voltage: \(V(0.2) = \frac{1}{10^{-5}} \times 0.2 \times \left(\frac{0.2 \times 10^{-3}}{2} + \frac{0.3683 \times 10^{-3}}{2} \right) \approx 5.6815 \times 10^{-2}\) V - Continue this process for other time intervals.
05

Plot voltage versus time

Once we have calculated the voltage at each time interval, we can plot the V(t) versus time graph using the obtained values. The graph will show how the voltage increases as the current passes through the capacitor.

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Key Concepts

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

Capacitor Voltage
A capacitor is a device that stores electrical energy in an electric field, and voltage across a capacitor is related to the charge it holds. When a capacitor initially holds no charge, the voltage across it as a function of time can be determined using an integral equation:
\[ V(t) = \frac{1}{C} \int_{0}^{t} i(t) \, dt \]
Here,
  • \(V(t)\) is the voltage at time \(t\),
  • \(C\) is the capacitance,
  • \(i(t)\) is the current flowing at time \(t\).
Capacitors are characterized by their capacitance \(C\), typically measured in farads. In this context, the given value of \(C\) is \(10^{-5}\) farads. The relationship shows that the voltage is proportional to the accumulated charge over time, reflecting the energy stored on the capacitor.
Trapezoidal Rule
When dealing with real-world data, perfect analytical integration is often not practical. This is where numerical methods such as the Trapezoidal Rule come into play. The Trapezoidal Rule helps approximate the integral of a function by dividing the area under the curve into trapezoids rather than perfect rectangles or other shapes.
The rule states that the integral from time \(0\) to \(t\) of \(i(t')\) can be approximated as:
\[ \int_{0}^{t} i(t') \, dt' \approx \Delta t \left( \frac{i_{0}}{2} + i_{1} + i_{2} + \dots + i_{n-1} + \frac{i_{n}}{2} \right) \]
  • Each \(i\) represents the current at a specific time interval.
  • \(\Delta t\) is the time interval between each measurement.
This approach is particularly useful when only discrete data points are available, allowing us to estimate the accumulated impact of current over a period of time.
Current Integration
Current integration is crucial for determining the voltage across a capacitor. Since the current is only presented at discrete intervals, you need to estimate the integral using numerical methods like the Trapezoidal Rule to capture how much current has flowed over time.
To integrate the current values given in the problem for voltage calculation, you:
  • Use the Trapezoidal Rule to calculate the integral at each time interval.
  • Multiply the result by the inverse of capacitance \(\frac{1}{C}\).
The process involves calculating a new voltage for each period based on how much charge was accumulated in that specific time step starting from \(t=0\) up to \(t=1.2\) seconds. This integration of current provides a cumulative effect of the incoming current on the voltage development.
Voltage Plotting
Once you've calculated the voltages for each time interval, plotting these values will visually represent how the voltage across the capacitor changes over time. This step involves the following:
  • Use a graphing tool or software to plot voltage values on the y-axis against time on the x-axis.
  • Each calculated voltage reflects how much energy is stored in the capacitor at its corresponding time interval.
The resulting plot showcases the relationship between current flow and the resulting voltage. It highlights how the voltage builds up gradually over time, offering insights into the capacitor's charging behavior. Such visualizations are pivotal in understanding dynamic systems in electronics, helping learners grasp how capacitive systems function in various applications.

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Most popular questions from this chapter

Fick s first diffusion law states that $$\text { Mass flux }=-D \frac{d c}{d x}$$ where mass flux \(=\) the quantity of mass that passes across a unit area per unit time \(\left(\mathrm{g} / \mathrm{cm}^{2} / \mathrm{s}\right), D=\) a diffusion coefficient \(\left(\mathrm{cm}^{2} / \mathrm{s}\right)\) \(c=\) concentration, and \(x=\) distance \((\mathrm{cm}) .\) An environmental engineer measures the following concentration of a pollutant in the sediments underlying a lake \((x=0\) at the sediment-water interface and increases downward): $$\begin{array}{l|ccc} x, \mathrm{cm} & 0 & 1 & 3 \\ \hline c, 10^{-6} \mathrm{g} / \mathrm{cm}^{3} & 0.06 & 0.32 & 0.6 \end{array}$$ Use the best numerical differentiation technique available to estimate the derivative at \(x=0 .\) Employ this estimate in conjunction with Eq. (P24.6) to compute the mass flux of pollutant out of the sediments and into the overlying waters \(\left(D=1.52 \times 10^{-6} \mathrm{cm}^{2} / \mathrm{s}\right)\) For a lake with \(3.6 \times 10^{6} \mathrm{m}^{2}\) of sediments, how much pollutant would be transported into the lake over a year's time?

The horizontal surface area \(A_{s}\left(\mathrm{m}^{2}\right)\) of a lake at a particular depth can be computed from volume by differentiation, $$A_{s}(z)=-\frac{d V}{d z}(z)$$ where \(V=\) volume \(\left(\mathrm{m}^{3}\right)\) and \(z=\) depth \((\mathrm{m})\) as measured from the surface down to the bottom. The average concentration of a substance that varies with depth \(\bar{c}\left(\mathrm{g} / \mathrm{m}^{3}\right)\) can be computed by integration $$\bar{c}=\frac{\int_{0}^{Z} c(z) A_{s}(z) d z}{\int_{0}^{Z} A_{s}(z) d z}$$ where \(Z=\) the total depth (m). Determine the average concentration based on the following data: $$\begin{array}{l|ccccc} z, m & 0 & 4 & 8 & 12 & 16 \\ \hline V, 10^{6} m^{3} & 9.8175 & 5.1051 & 1.9635 & 0.3927 & 0.0000 \\ \hline c, g / m^{3} & 10.2 & 8.5 & 7.4 & 5.2 & 4.1 \end{array}$$

Suppose that the current through a resistor is described by the function $$i(t)=(60-t)^{2}+(60-t) \sin (\sqrt{t})$$ and the resistance is a function of the current, $$R=10 i+2 i^{2 / 3}$$

Using the following data, calculate the work done by stretching a spring that has a spring constant of \(k=300 \mathrm{N} / \mathrm{m}\) to \(x=0.35 \mathrm{m}\) $$\begin{array}{l|llllllll} F, 10^{3} \mathrm{N} & 0 & 0.01 & 0.028 & 0.046 & 0.063 & 0.082 & 0.11 & 0.13 \\\ \hline x, \mathrm{m} & 0 & 0.05 & 0.10 & 0.15 & 0.20 & 0.25 & 0.30 & 0.35 \end{array}$$

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