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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 given times using formulas, plot for visualization.

Step by step solution

01

Understand the Problem

We have a capacitor and resistor in series connected to a steady voltage source. Our task is to compute the charge and current at various times after the switch connecting them is closed.
02

Identify Key Formulas

The formula for the charge on a capacitor as it charges over time is given by: \[ q(t) = Q(1 - e^{-t/(RC)}) \]where \( Q = C \times V \) is the maximum charge (\( C \) is capacitance and \( V \) is the voltage), \( R \) is the resistance, and \( t \) is time. The current as a function of time is given by:\[ I(t) = \frac{V}{R}e^{-t/(RC)} \]
03

Calculate Maximum Charge

Calculate the maximum charge \( Q \) that the capacitor can hold using the formula:\[ Q = C \times V \]Given \( C = 12.4 \times 10^{-6} \ \mathrm{F} \) and \( V = 60.0 \ \mathrm{V} \), we find:\[ Q = 12.4 \times 10^{-6} \times 60.0 = 0.744 \times 10^{-3} \ \mathrm{C} \].
04

Compute Charge at Different Times

Use the formula for charge:\[ q(t) = Q(1 - e^{-t/(RC)}) \]- At \( t = 0 \): \[ q(0) = 0.744 \times 10^{-3} \times (1 - e^{0}) = 0 \ \mathrm{C} \]- At \( t = 5 \ \mathrm{s} \): substitute \( t \) with 5- At \( t = 10 \ \mathrm{s} \): substitute \( t \) with 10- At \( t = 20 \ \mathrm{s} \): substitute \( t \) with 20- At \( t = 100 \ \mathrm{s} \): substitute \( t \) with 100And calculate using \( R = 0.895 \times 10^{6} \ \Omega \), \( C = 12.4 \times 10^{-6} \ \mathrm{F} \).
05

Compute Current at Different Times

Use the formula for current:\[ I(t) = \frac{V}{R}e^{-t/(RC)} \]- At \( t = 0 \): \[ I(0) = \frac{60.0}{0.895 \times 10^{6}} \times e^{0} \]- At \( t = 5 \ \mathrm{s} \): substitute \( t \) with 5- At \( t = 10 \ \mathrm{s} \): substitute \( t \) with 10- At \( t = 20 \ \mathrm{s} \): substitute \( t \) with 20- At \( t = 100 \ \mathrm{s} \): substitute \( t \) with 100And calculate using previous values.
06

Graph the Results

Plot charge \( q(t) \) and current \( I(t) \) against time from 0 to 20 seconds. Observe how the charge increases, approaching maximum, and the current decreases to zero as time goes by.

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Key Concepts

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

Capacitor Charging
When a capacitor charges, it stores electrical energy by accumulating electrons on its plates. This situation occurs when a circuit connects a capacitor and a resistor to a power source, like a battery, allowing a flow of current. As the capacitor charges, the voltage across its plates grows until it equals the battery's voltage. Initially, the charging current is at its maximum because the potential difference between the power source and the capacitor is greatest. Over time, this current gradually decreases as the voltage across the capacitor approaches the power source voltage. The formula that describes capacitor charging is given by
  • \( q(t) = Q(1 - e^{-t/(RC)}) \)
where \( q(t) \) is the charge at a time \( t \), \( Q \) is the maximum charge, \( R \) is the resistance, and \( C \) is the capacitance. The exponential term \( e^{-t/(RC)} \) explains how the charge grows progressively to its maximum. This process of charging resembles filling a cup with a fixed flow, slowing down as it nears full capacity.
Time Constant
A crucial aspect in understanding capacitor circuits is the time constant, usually denoted by \( \tau \) (tau). This parameter determines how fast or slow the capacitor charges or discharges. The time constant \( \tau \) is calculated by multiplying the resistance \( R \) by the capacitance \( C \):
  • \( \tau = RC \)
In our exercise, with a resistor of \(0.895\,\text{M}\Omega\) and a capacitor of \(12.4\,\mu \text{F}\), the time constant is
  • \( \tau = 0.895 \times 10^6 \times 12.4 \times 10^{-6} = 11.098\,\text{s} \).
This value indicates that after approximately 11 seconds, the capacitor's charge reaches about 63.2% of its maximal value. The time constant is a handy tool for predicting how quickly a capacitor will charge or discharge in circuits, often used for designing and analyzing electronic filters and timers.
Exponential Growth and Decay
In the context of RC circuits, exponential growth and decay describe how quantities like charge and current evolve over time. For capacitor charging, the growth of charge on the capacitor follows an exponential pattern, where it gradually approaches a maximum value. The term \(1 - e^{-t/(RC)}\) in the charge formula reflects this exponential growth. **Exponential Growth**:
  • Starts rapidly and then slows down.
  • Charge builds up quickly at first as the potential difference is largest.
Conversely, when discharging a capacitor, the charge and current show exponential decay, meaning they decrease rapidly at first, then slowly over time. **Exponential Decay**:
  • Occurs in opposite direction, rapidly decreasing.
  • Describes the drop in current as the circuit stabilizes.
Understanding these exponential patterns is essential because they reveal the natural behavior of electrical circuits in responding to changes in voltage and current over time, a concept valuable in electronics and signal processing.

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Most popular questions from this chapter

An \(R\) -C circuit has a time constant \(R C\) (a) If the circuit is discharging, how long will it take for its stored energy to be reduced to 1/e of its initial value? (b) If it is charging, how long will it take for the stored energy to reach 1\(/ e\) of its maximum value?

Cp The heating element of an electric stove consists of a heater wire embedded within an electrically insulating material, which in turn is inside a metal casing. The heater wire has a resistance of 20\(\Omega\) at room temperature \(\left(23.0^{\circ} \mathrm{C}\right)\) and a temperature coefficient of resistivity \(\alpha=2.8 \times 10^{-3}\left(\mathrm{C}^{0}\right)^{-1} .\) The heating element operates from a 120 \(\mathrm{-V}\) line. (a) When the heating element is first turned on, what current does it draw and what electrical power does it dissipate? (b) When the heating element has reached an operating temperature of \(280^{\circ} \mathrm{C}\left(536^{\circ} \mathrm{F}\right),\) what current does it draw and what electrical power does it dissipate?

A Three-Way Light Bulb. A three-way light bulb has three brightness settings (low, medium, and high) but only two filaments. (a) A particular three-way light bulb connected across a 120 -V line can dissipate \(60 \mathrm{W}, 120 \mathrm{W},\) or 180 \(\mathrm{W}\) . Describe how the two filaments are arranged in the bulb, and calculate the resistance of each filament. (b) Suppose the filament with the higher resistance burns out. How much power will the bulb dissipate on each of the three brightness settings? What will be the brightness (low, medium, or high) on each setting? (c) Repeat part (b) for the situation in which the filament with the lower resistance burns out.

A \(12.0-\mu \mathrm{F}\) capacitor is charged to a potential of 50.0 \(\mathrm{V}\) and then discharged through a 175 - \(\Omega\) resistor. How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy?

Two identical \(3.00-\Omega\) wires are laid side by side and soldered together so they touch each other for half of their lengths. What is the equivalent resistance of this combination?

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