/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 An initially uncharged capacitor... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The energy stored in the capacitor and dissipated by the resistor are each half of the energy supplied by the emf device.

Step by step solution

01

Understanding the Setup

We have a capacitor with capacitance \(C\) initially uncharged. It is being charged by an emf source \(\mathscr{E}\) with a resistor \(R\) in series. The charging process follows an exponential profile governed by the circuit's time constant \(\tau = R \cdot C\).
02

Calculating the Final Energy Stored in the Capacitor

The final charge \(Q_f\) on the capacitor when fully charged is given by \(Q_f = C \cdot \mathscr{E}\). The energy \(U_f\) stored in the capacitor is \(U_f = \frac{1}{2} C \cdot \mathscr{E}^2\).
03

Calculating the Total Energy Supplied by the EMF

The emf source supplies energy \(W_s\) given by \(W_s = \mathscr{E} \cdot Q_f = \mathscr{E} \cdot C \cdot \mathscr{E} = C \cdot \mathscr{E}^2\).
04

Relationship Between Energy Supplied and Stored

Compare the energy stored by the capacitor \(U_f = \frac{1}{2} C \cdot \mathscr{E}^2\) to the energy supplied by the emf device \(W_s = C \cdot \mathscr{E}^2\). We see \(U_f = \frac{1}{2} W_s\), or half of the supplied energy is stored in the capacitor.
05

Calculating Thermal Energy Dissipated in Resistor

The power dissipated in the resistor is \(P(t) = i(t)^2 \times R\), where \(i(t) = \frac{\mathscr{E}}{R} e^{-t/RC}\). Integrate \(P(t)\) over time to find energy dissipated:\[\text{Thermal Energy} = \int_0^\infty \left(\frac{\mathscr{E}}{R} e^{-t/RC}\right)^2 R \, dt\]Calculate this integral to confirm thermal energy is \(\frac{1}{2} C \cdot \mathscr{E}^2\), proving it equals the other half of the energy supplied by the emf device.

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Key Concepts

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

Charging Process
The charging process of a capacitor involves making it store energy by connecting it to a power source or an electromotive force (emf) through a circuit. Initially, the capacitor has no charge. Once connected, the capacitor begins to accumulate charge because of the voltage source, giving rise to a current in the circuit.
Saturating the capacitor, or fully charging it, requires the full battery voltage to be applied across it. As the charge builds up on the capacitor plates, the voltage across its terminals also increases to equal the source voltage. The flow of current diminishes over time, following an exponential decay, until it reaches zero. At this point, the capacitor is fully charged, and no more current flows in the circuit.
Energy Dissipation
When charging a capacitor through a resistor, not all supplied energy is stored. Some of the energy dissipates as thermal energy, primarily by the resistor, during the charging process. This phenomenon often leads to inefficiencies in electrical circuits.
In simple terms, while charging, the energy from the emf device is divided equally: half is stored in the capacitor, while the other half dissipates as heat in the resistor. During the charging phase, the resistor converts some of the electrical energy into heat due to the current flowing through it. This conversion is a significant feature of any real-world circuit utilizing resistors.
Resistor in Series
In circuits, a resistor in series with a capacitor functions as a crucial component to control the charging rate of the capacitor. This is because a resistor limits the amount of current entering the capacitor, essentially dictating how fast the capacitor charges.
The presence of this resistor leads to a characteristic charging curve described by an exponential function. Without such resistance, the capacitor would charge almost instantaneously, which might not be desired in many practical applications. Additionally, this resistor helps prevent potential damage by regulating the electrical current and ensuring safe operation.
Exponential Profile
An exponential profile characterizes the rate at which a capacitor charges or discharges. As the charge accumulates, the process begins rapidly and then slows down exponentially over time.
Mathematically, this relationship is depicted using the exponential function: the current decreases exponentially, \[ i(t) = \frac{\mathscr{E}}{R} e^{-t/RC} \]where \(i(t)\) is the instantaneous current, \(\mathscr{E}\) is the emf, \(R\) is the resistance, and \(RC\) is the time constant. The time constant \(\tau = RC\) is crucial as it determines the speed of the charging and discharging processes. When \(t = RC\), the capacitor is about 63% charged. As time proceeds to infinity, the capacitor asymptotically approaches its full charge.

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Most popular questions from this chapter

Two identical batteries of emf \(\mathscr{h}=\) \(12.0 \mathrm{~V}\) and internal resistance \(r=0.200 \Omega\) are to be connected to an external resistance \(R\), either in parallel (Fig. \(27-50\) ) or in series (Fig. 27-51). If \(R=2.00 r\), what is the current \(i\) in the external resistance in the (a) parallel and (b) series arrangements? (c) For which arrangement is \(i\) greater? If \(R=\) \(r / 2.00\), what is \(i\) in the external resistance in the (d) parallel arrangement and (e) series arrangement? (f) For which arrangement is \(i\) greater now?

A \(1.0 \mu \mathrm{F}\) capacitor with an initial stored energy of \(0.50 \mathrm{~J}\) is discharged through a \(1.0 \mathrm{M} \Omega\) resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time \(t,(\mathrm{c})\) the potential difference \(V_{C}\) across the capacitor, (d) the potential difference \(V_{R}\) across the resistor, and (e) the rate at which thermal energy is produced in the resistor.

Four \(18.0 \Omega\) resistors are connected in parallel across a \(25.0 \mathrm{~V}\) ideal battery. What is the current through the battery?

An automobile gasoline gauge is shown schematically in Fig. \(27-74\). The indicator (on the dashboard) has a resistance of \(10 \Omega\). The tank unit is a float connected to a variable resistor whose resistance varies linearly with the volume of gasoline. The resistance is \(140 \Omega\) when the tank is empty and \(20 \Omega\) when the tank is full. Find the current in the circuit when the tank is (a) empty, (b) half-full, and (c) full. Treat the battery as ideal.

A car battery with a \(12 \mathrm{~V}\) emf and an internal resistance of \(0.040 \Omega\) is being charged with a current of \(50 \mathrm{~A}\). What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{r}\) of energy dissipation inside the battery, and (c) the rate \(P_{\text {cmt }}\) of energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) \(V\) and (e) \(P_{r} ?\)

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