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A 12.4 \(\mu \mathrm{F}\) capacitor is connected through a 0.895 \(\mathrm{M\Omega}\) resistor to a constant potential difference of 60.0 \(\mathrm{V}\) . (a) Compute the charge on the capacitor at the following times after the connections are made: \(0,5.0 \mathrm{s}, 10.0 \mathrm{s}, 20.0 \mathrm{s},\) and 100.0 \(\mathrm{s}\) . (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and 20 \(\mathrm{s}\) .

Short Answer

Expert verified
Calculate charge and current at times using exponential growth formulas and graph the results.

Step by step solution

01

Understand the problem

We need to calculate the charge on the capacitor and the current flowing through the resistor at different time intervals after the capacitor starts charging. The equations governing these variables for a charging capacitor are based on exponential growth functions.
02

Determine the time constant

The time constant of an RC circuit is given by \( \tau = R \cdot C \). Given that \( R = 0.895 \, \mathrm{M\Omega} = 0.895 \times 10^6 \, \Omega \) and \( C = 12.4 \, \mu\mathrm{F} = 12.4 \times 10^{-6} \, \mathrm{F} \), calculate \( \tau \). \[ \tau = 0.895 \times 10^6 \cdot 12.4 \times 10^{-6} = 11.098 \, \mathrm{s} \]
03

Use the formula for charge on the capacitor

The charge \( Q \) on a capacitor at any time \( t \) during the charging process is given by \( Q(t) = C \cdot V (1 - e^{-t/\tau}) \), where \( C \) is the capacitance, \( V \) is the potential difference, and \( \tau \) is the time constant. Our goal is to calculate \( Q \) at \( t = 0, 5, 10, 20, \) and \( 100 \; \mathrm{s} \).
04

Calculate charge at given times

Calculate \( Q(t) \) using the formula for each specific time:- At \( t = 0 \) s: \( Q(0) = 12.4 \times 10^{-6} \times 60 \times (1 - e^{0}) = 0 \; \mathrm{C} \).- At \( t = 5 \) s: \( Q(5) = 12.4 \times 10^{-6} \times 60 \times (1 - e^{-5/11.098}) \approx 3.27 \; \mathrm{C} \).- At \( t = 10 \) s: \( Q(10) = 12.4 \times 10^{-6} \times 60 \times (1 - e^{-10/11.098}) \approx 5.31 \; \mathrm{C} \).- At \( t = 20 \) s: \( Q(20) = 12.4 \times 10^{-6} \times 60 \times (1 - e^{-20/11.098}) \approx 7.77 \; \mathrm{C} \).- At \( t = 100 \) s: \( Q(100) = 12.4 \times 10^{-6} \times 60 \times (1 - e^{-100/11.098}) \approx 7.91 \; \mathrm{C} \).
05

Use the formula for charging current

The current \( I \) at any time \( t \) is given by the derivative of charge, which is \( I(t) = \frac{V}{R} \, e^{-t/\tau} \). Compute \( I(t) \) at each given time value.
06

Calculate current at given times

Compute \( I(t) \) using the formula for each specific time:- At \( t = 0 \) s: \( I(0) = \frac{60}{0.895 \times 10^6} \times 1 \approx 6.71 \times 10^{-5} \; \mathrm{A} \).- At \( t = 5 \) s: \( I(5) = \frac{60}{0.895 \times 10^6} e^{-5/11.098} \approx 4.02 \times 10^{-5} \; \mathrm{A} \).- At \( t = 10 \) s: \( I(10) = \frac{60}{0.895 \times 10^6} e^{-10/11.098} \approx 2.41 \times 10^{-5} \; \mathrm{A} \).- At \( t = 20 \) s: \( I(20) = \frac{60}{0.895 \times 10^6} e^{-20/11.098} \approx 8.63 \times 10^{-6} \; \mathrm{A} \).- At \( t = 100 \) s: \( I(100) = \frac{60}{0.895 \times 10^6} e^{-100/11.098} \approx 4.80 \times 10^{-7} \; \mathrm{A} \).
07

Graph the results

Using the results from Step 4 and Step 6, plot the graphs of \( Q(t) \) versus \( t \) and \( I(t) \) versus \( t \) for times between 0 and 20 seconds.Make sure the charge graph starts from zero and approaches the maximum charge asymptotically. The current graph should start from a maximum at \( t = 0 \) and decrease exponentially towards zero.

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Key Concepts

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

RC circuit
An RC circuit is an electric circuit composed of a resistor (R) and a capacitor (C) connected in series or parallel with a voltage source. In our given problem, the circuit consists of a capacitor with capacitance of 12.4 microfarads (\( \mu F \)) and a resistor with resistance of 0.895 megaohms (\( M\Omega \)), connected to a 60-volt power supply.

These circuits are crucial in many electronic applications due to their ability to filter signals, store energy, and manage the timing of electronic processes. The key characteristic of an RC circuit is its ability to charge and discharge a capacitor over time.

When you connect the power supply, the capacitor starts charging, and the process of charging is dictated by the properties of the capacitor and resistor. During this, the voltage across the capacitor gradually rises until it equals the voltage of the power supply. This charging process is not instantaneous, and it follows an exponential pattern which we will explore in the next sections.
time constant
The time constant, denoted by \( \tau \), is a crucial factor in an RC circuit. It quantifies how quickly the circuit responds to changes. Specifically, in an RC circuit, the time constant is defined as the product of the resistance and the capacitance:\[ \tau = R \times C \]

In our specific example, we calculate:
  • Resistance \( R = 0.895 \times 10^6 \Omega \)
  • Capacitance \( C = 12.4 \times 10^{-6} F \)
  • Resulting in a time constant \( \tau = 11.098 \, \mathrm{s} \)
This result tells us how long it takes for the capacitor to reach approximately 63.2% of its full charge. Thus, predicting how the voltage and current evolve over time. The larger the time constant, the slower the circuit charges.

The time constant makes it easier to determine the behavior of the circuit without needing to solve complex equations all the time. Understanding \( \tau \) helps us know when the capacitor is almost fully charged or discharged.
exponential growth
Exponential growth describes the way quantities increase at a consistent rate over time. In the context of an RC circuit, this growth pattern is seen in how the capacitor charges over time. As the charging begins, the charge on the capacitor (\( Q(t) \)) increases exponentially according to the formula:

\[ Q(t) = C \cdot V (1 - e^{-t/\tau}) \]

Here, \(e\) is the base of the natural logarithm (approximately equal to 2.718), \(V\) is the potential difference, and \(\tau\) is the time constant.

This formula indicates that:
  • At \( t = 0 \), the charge is 0 because \( 1 - e^{0} = 0 \).
  • As time progresses, \( e^{-t/\tau} \) decreases, causing \( Q(t) \) to rise.
  • Eventually, as \( t \) becomes very large, \( Q(t) \) approaches \( C \cdot V \). This is the maximum charge, fully achieved theoretically after an infinite amount of time.
The growth is not linear, meaning that the charge doesn't accumulate evenly over time. Instead, it slows down as the capacitor nears its full capacity. This non-linear behavior is typical of many real-world phenomena beyond just electronics.

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Most popular questions from this chapter

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