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A \(3.00 \mathrm{M} \Omega\) resistor and a \(1.00 \mu \mathrm{F}\) capacitor are connected in series with an ideal battery of emf \(\mathscr{E}=4.00 \mathrm{~V}\). At \(1.00 \mathrm{~s}\) after the connection is made, what is the rate at which (a) the charge of the capacitor is increasing, (b) energy is being stored in the capacitor, (c) thermal energy is appearing in the resistor, and (d) energy is being delivered by the battery?

Short Answer

Expert verified
(a) 1.03 µC/s, (b) 1.06 µW, (c) 3.18 µW, (d) 4.24 µW.

Step by step solution

01

Calculate initial current using Ohm's law

For an RC circuit, the initial current through the circuit immediately after the battery is connected is determined by Ohm's law: \( I = \frac{\mathscr{E}}{R} \). Here, \( \mathscr{E} = 4.00\,\mathrm{V} \) and \( R = 3.00\,\mathrm{M}\Omega = 3.00 \times 10^6\,\Omega \). So, the initial current is:\[ I_0 = \frac{4.00}{3.00 \times 10^6} = 1.33 \times 10^{-6}\,\mathrm{A} \].
02

Calculate the current at t = 1 s using the exponential decay formula

The current \( I(t) \) in the RC circuit as a function of time is given by the formula: \( I(t) = I_0 e^{-t/RC} \). The time constant \( \tau = RC \) is \( 3.00 \times 10^6\,\Omega \times 1.00 \times 10^{-6}\, \mathrm{F} = 3.00\,\mathrm{s} \). Therefore:\[ I(1) = 1.33 \times 10^{-6} e^{-1/3} \approx 1.03 \times 10^{-6}\,\mathrm{A} \].
03

Find charge increasing rate on the capacitor

The rate at which the charge \( Q \) on the capacitor is increasing is equal to the current through the circuit at \( t = 1\,\mathrm{s} \), which is \( dQ/dt = I(1) = 1.03 \times 10^{-6}\,\mathrm{A} \) or \( 1.03\,\mu\mathrm{C/s} \).
04

Calculate power stored in the capacitor

The power being stored in the capacitor is given by \( P_C = I^2(t) R_{ ext{eff}} \), where \( R_{\text{eff}} = R/2 \) because power is split evenly between resistor and capacitor. Thus:\[ P_C = (1.03 \times 10^{-6})^2 \times 1.5 \times 10^6 = 1.06 \times 10^{-6}\,\mathrm{W} \].
05

Calculate thermal energy appearing in the resistor

Thermal power dissipated in the resistor is given by \( P_R = I^2(t) R \). Therefore:\[ P_R = (1.03 \times 10^{-6})^2 \times 3.00 \times 10^6 = 3.18 \times 10^{-6}\,\mathrm{W} \].
06

Calculate total energy delivery by the battery

The energy delivered by the battery is the sum of energy stored in the capacitor and energy dissipated as heat in the resistor. Thus:\[ P_B = P_C + P_R = 1.06 \times 10^{-6} + 3.18 \times 10^{-6} = 4.24 \times 10^{-6}\,\mathrm{W} \].

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in electrical circuits, relating the current flowing through a conductor to the voltage across it and the resistance it encounters. It is expressed as \( I = \frac{V}{R} \), where \( I \) is the current in amperes, \( V \) is the voltage in volts, and \( R \) is the resistance in ohms. This law is crucial for analyzing circuits because it allows us to determine how much current will flow for a given voltage applied, considering the resistance.In our original exercise, Ohm's Law helped us find the initial current in the RC circuit. With the battery having an electromotive force (EMF) of 4.00 V and a resistor of 3.00 MΩ, we calculated the initial current right after the circuit was completed. This calculation provides a starting point for further analysis of the circuit behavior over time.
Exponential Decay in Circuits
Exponential decay is a concept that describes how quantities decrease over time at a rate proportional to their current value. In the context of electrical circuits, it appears when current and voltage across capacitors change as they charge or discharge.For an RC Circuit, the time-dependent behavior follows an exponential equation \( I(t) = I_0 e^{-t/RC} \), where \( I(t) \) is the current at time \( t \), \( I_0 \) is the initial current, and \( \tau = RC \) is the time constant. The time constant \( \tau \) indicates how quickly the circuit will reach steady state, with larger \( \tau \) values indicating slower changes.In the given exercise, this formula was used to determine the current at one second after making the connection. The exponential factor \( e^{-t/RC} \) reflects the decrease in current as the capacitor charges up, showing how energy distribution dynamically evolves.
Energy Storage in Capacitors
Capacitors are vital components in circuits, capable of storing and releasing electrical energy. Energy storage in a capacitor occurs as it accumulates charge, and the energy \( U \) stored is given by the formula \( U = \frac{1}{2}CV^2 \), where \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.For the exercise, the stored energy reflects how the charge increases over time, powered by the battery. This energy can fluctuate based on the current flow and the effective resistance. Calculating power stored in the capacitor, as done in the step-by-step solution, provides insight into the capacitor's role in the overall circuit energy dynamics.Understanding this process is crucial for interpreting how capacitor performance affects circuit behavior, particularly in timing applications and energy supply smoothing.
Thermal Energy in Resistors
Resistors dissipate electrical energy as thermal energy (or heat), a process central to their function of controlling current flow in circuits. The power \( P \) dissipated as heat in a resistor is calculated as \( P = I^2 R \), where \( I \) is current through the resistor and \( R \) is its resistance.Thermal energy loss is inherent in any electrical circuit and represents efficiency losses, often perceived as waste heat. In the context of the given problem, we calculated the thermal power appearing in the resistor at a specific moment. This calculation highlights how energy is transformed within the circuit, with some energy converted to heat, which must be considered in thermal management tasks.This concept is critical in designing circuits to ensure components operate safely within their thermal limits, and that sufficient energy is delivered to achieve desired outcomes.

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Most popular questions from this chapter

The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other) \(2.0 \mu \mathrm{F}\) capacitor drops to one-fourth its initial value in \(2.0 \mathrm{~s}\). What is the equivalent resistance between the capacitor plates?

A \(5.0 \mathrm{~A}\) current is set up in a circuit for \(6.0\) min by a rechargeable battery with a \(6.0 \mathrm{~V}\) emf. By how much is the chemical energy of the battery reduced?

Suppose that, while you are sitting in a chair, charge separation between your clothing and the chair puts you at a potential of \(200 \mathrm{~V}\), with the capacitance between you and the chair at \(150 \mathrm{pF}\). When you stand up, the increased separation between your body and the chair decreases the capacitance to \(10 \mathrm{pF}\). (a) What then is the potential of your body? That potential is reduced over time, as the charge on you drains through your body and shoes (you are a capacitor discharging through a resistance). Assume that the resistance along that route is 300 \(\mathrm{G} \Omega\). If you touch an electrical component while your potential is greater than \(100 \mathrm{~V}\), you could ruin the component. (b) How long must you wait until your potential reaches the safe level of \(100 \mathrm{~V} ?\) If you wear a conducting wrist strap that is connected to ground, your potential does not increase as much when you stand up; you also discharge more rapidly because the resistance through the grounding connection is much less than through your body and shoes. (c) Suppose that when you stand up, your potential is \(1400 \mathrm{~V}\) and the chair-to-you capacitance is \(10 \mathrm{pF}\). What resistance in that wrist-strap grounding connection will allow you to discharge to 100 \(\mathrm{V}\) in \(0.30 \mathrm{~s}\), which is less time than you would need to reach for, say, your computer?

A capacitor with initial charge \(q_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

An initially uncharged capacitor \(C\) is fully charged by a device of constant emf \(\mathscr{E}\) connected in series with a resistor \(R\). (a) Show that the final energy stored in the capacitor is half the energy supplied by the emf device. (b) By direct integration of \(i^{2} R\) over the charging time, show that the thermal encrgy dissipated by the resistor is also half the energy supplied by the emf device.

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