/*! 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 84 A resistor with \(R=850 \Omega\)... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 84

A resistor with \(R=850 \Omega\) is connected to the plates of a charged capacitor with capacitance \(C=4.62 \mu F\) . Just before the connection is made. the charge on the capacitor is 8.10 \(\mathrm{mC}\) .connection is made, the charge on the capacitor is 8.10 \(\mathrm{mC}\) . (a) What is the energy initially stored in the capacitor? (b) What is the electrical power dissipated in the resistor just after the connec- tion is made? (c) What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part (a)?

Short Answer

Expert verified
(a) Initial energy is approximately 7.10 J. (b) Initial power is approximately 77.0 W. (c) Power at half-energy is approximately 38.5 W.

Step by step solution

01

Calculate Initial Energy in Capacitor

The energy stored in a capacitor is given by the formula \( U = \frac{1}{2} C V^2 \). Since we know the charge \( Q = 8.10 \, \text{mC} = 8.10 \times 10^{-3} \, \text{C} \), and we can find the voltage \( V \) using the formula \( V = \frac{Q}{C} \). First, convert \( C = 4.62 \, \mu F = 4.62 \times 10^{-6} \, \text{F} \). Calculate \( V \) and then substitute in the energy formula to find \( U \).
02

Convert Charge and Capacitance

Convert the given charge to Coulombs: \( 8.10 \, \text{mC} = 8.10 \times 10^{-3} \, \text{C} \). Convert the capacitance to Farads: \( 4.62 \, \mu F = 4.62 \times 10^{-6} \, \text{F} \).
03

Compute Voltage Across Capacitor

Find the voltage \( V \) across the capacitor using \( V = \frac{Q}{C} = \frac{8.10 \times 10^{-3}}{4.62 \times 10^{-6}} \).
04

Calculate Initial Energy

Substitute \( V \) into the equation \( U = \frac{1}{2} C V^2 \) to find the initial energy stored in the capacitor. Calculate \( U \).
05

Determine Initial Power Dissipation

The power dissipated in the resistor just after the connection is made is given by \( P = \frac{V^2}{R} \). Use the voltage from Step 3 to find the initial power dissipation.
06

Calculate Energy When Reduced to Half

When energy is half: \( U_{\text{half}} = \frac{1}{2} U \). Re-solve the energy formula for \( V_{\text{new}} \) when \( U = \frac{1}{2} U_\text{initial} = \frac{\left( 1/2 \right) (1/2) C V^2}{1/2} \). Find the new voltage.
07

Determine Electrical Power at Half Energy

Calculate the power at half-energy using \( P = \frac{V_{\text{new}}^2}{R} \). Substitute the half voltage from Step 6 to find the power dissipation at this point.

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Key Concepts

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

Capacitor Energy
Capacitors store energy in an electric field between their plates. The amount of energy a capacitor can hold is dictated by its capacitance and the voltage across it. To calculate the energy, we use the formula \( U = \frac{1}{2} C V^2 \).
To find the voltage \( V \) across the capacitor, given a known charge \( Q \), we use the relation \( V = \frac{Q}{C} \). Here, \( Q \) is the charge, and \( C \) is the capacitance.
Once we calculate \( V \), we can plug it into the energy formula to find the energy \( U \). This process allows us to determine how much electrical energy is stored initially, just before the connected circuit begins to discharge the capacitor.
Resistor Power Dissipation
When a charged capacitor is connected to a resistor, it begins to release its stored energy through the resistor. This process dissipates power in the form of heat. The power dissipated by a resistor is given by \( P = \frac{V^2}{R} \), where \( V \) is the voltage across the resistor, and \( R \) is the resistance.
As soon as the capacitor begins to discharge, the initial voltage across the resistor is the same as that across the capacitor. Therefore, the initial power dissipation is at its maximum value just when the connection is made.
This concept is vital in circuits as it helps us understand where and how energy is lost or converted within electronic components.
RC Circuit Analysis
An RC circuit is composed of a Resistor (R) and a Capacitor (C) connected in series. The behavior of this circuit is characterized by how the voltage and current change over time. The discharge of an initially charged capacitor through the resistor follows an exponential decay pattern.
The time constant \( \tau \) of an RC circuit, which defines the rate of this change, is given by \( \tau = RC \). This time constant helps predict how quickly the capacitor will discharge to \( \frac{1}{e} \) of its initial charge.
Understanding the dynamics of an RC circuit is essential for analyzing how quickly a capacitor can release its energy and how quickly it can be recharged.
Charge and Voltage Relation
In electrical circuits, the relationship between charge and voltage is fundamental, especially in capacitors. The charge \( Q \) stored in a capacitor is directly proportional to the voltage \( V \) across it, as given by the equation \( Q = CV \).
This relationship signifies that with an increase in voltage, the amount of charge stored also increases, assuming the capacitance remains constant. Given the capacitance and charge, one can determine the voltage across the capacitor, which is particularly useful in design and analysis of circuits.
This charge-voltage relationship is not only central to understanding capacitors but also forms a basis for deeper exploration into various electronic components and circuit behavior.

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Most popular questions from this chapter

A resistor and a capacitor areconnected in series to an emf source. The time constant for the circuit is 0.870 \(\mathrm{s}\) . (a) A second capacitor, identical to the first, is added in series. What is the time constant for this new circuit? (b) In the original circuit a second capacitor, identical to the first, is con- nected in parallel with the first capacitor. What is the time constant for this new circuit?

The heating element of an clectric dryer is rated at 4.1 \(\mathrm{kW}\) when connected to a \(240-\mathrm{V}\) line. (a) What is the current in the heating element? Is 12 -gauge wire large enough to supply this current? (b) What is the resistance of the dryer's heating element at its operating temperature? (c) At 11 cents per \(\mathrm{kWh}\) , how much does it cost per hour to operate the dryer?

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 \(\alpha=2.8 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\) . The heating element oper- ates 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 dis- sipate? (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. \(2.36-\mu F\) capacitor that is initially uncharged is connected in series with a \(4.26-\Omega\) resistor and an emf source with \(\mathcal{E}=120 \mathrm{V}\) and negligible intermal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value.

A \(1500-\) W electric heater is plugged into the outlet of a 120. V circuit that has a \(20-A\) circuit breaker. You plug an electric hair dryer into the same outlet. The hair dryer has power settings of \(600 \mathrm{W}, 900 \mathrm{W}, 1200 \mathrm{W},\) and 1500 \(\mathrm{W}\) . You start with the hair dryer on the \(600-\mathrm{W}\) setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?
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