91Ó°ÊÓ

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.

Step by step solution

01

Just After Connection - Calculate Initial Current

When the switch is first closed, the capacitor is uncharged, so its initial voltage is zero. The entire voltage from the source appears across the resistor.Using Ohm's Law: \[ I = \frac{\mathcal{E}}{R} = \frac{120\, \text{V}}{4.26\, \Omega} \approx 28.17\, \text{A} \]

02

Just After Connection - Calculate Power Dissipated in Resistor

The power dissipated in the resistor can be calculated using the formula: \[ P_R = I^2 R = (28.17 \text{ A})^2 \times 4.26 \Omega \approx 3380.29 \text{ W} \]

03

Just After Connection - Calculate Rate of Energy Storage in Capacitor

Since the capacitor is initially uncharged, the energy storage rate is zero just after the switch is closed. \[ P_C = 0 \]

04

Just After Connection - Calculate Power Output of Source

The power output of the source is the same as the power dissipated in the resistor plus the power stored in the capacitor:\[ P_{source} = P_R + P_C = 3380.29 \text{ W} + 0 \text{ W} = 3380.29 \text{ W} \]

05

Long Time After Connection - Calculate Steady-State Current

When a long time has passed, the capacitor is fully charged, and no current flows.\[ I = 0 \]

06

Long Time After Connection - Calculate Power Dissipated in Resistor

Since the current is zero, the power dissipated in the resistor is zero.\[ P_R = 0 \]

07

Long Time After Connection - Calculate Rate of Energy Storage in Capacitor

The capacitor is fully charged, so energy storage rate in the capacitor is zero.\[ P_C = 0 \]

08

Long Time After Connection - Calculate Power Output of Source

Since everything has reached steady state, the power output of the source is zero.\[ P_{source} = 0 \]

09

When Capacitor is Half Charged - Solve for Time Constant

The charge on the capacitor increases according to the equation: \[ q(t) = C \mathcal{E} (1 - e^{-t/RC}) \]When the charge is half its maximum value, \[ q(t) = \frac{C \mathcal{E}}{2} \]Solving for \( t \) gives:\[ t = RC \ln(2) \]

10

When Capacitor is Half Charged - Calculate Current

At this instant, the current is:\[ I = \frac{\mathcal{E}}{R} e^{-t/RC} \]Using \( t = RC \ln(2) \), \[ I = \frac{120}{4.26} \times \frac{1}{2} \approx 14.085 \text{ A} \]

11

When Capacitor is Half Charged - Calculate Power Dissipated in Resistor

The power dissipated is:\[ P_R = I^2 R = (14.085 \text{ A})^2 \times 4.26 \Omega \approx 847.57 \text{ W} \]

12

When Capacitor is Half Charged - Calculate Power Stored in Capacitor

At this instant, since we determined the charge is half its final value, the rate change of energy storage is given as the rest difference:\[ P_{C} = P_{Source} - P_{R} = \text{initial power} - \text{power dissipated now} = 1690.145 \text{ W} - 847.57 \text{ W}\approx 847.57 \text{ W}\]

13

When Capacitor is Half Charged - Calculate Power Output of Source

The power output of the source is the sum of the power in the resistor and the power in the capacitor:\[ P_{source} = P_R + P_C = 847.57 \text{ W} + 847.57 \text{ W} \approx 1695.14 \text{ W} \]

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Key Concepts

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

Capacitor Charging and Discharging

Understanding how capacitors charge and discharge in an RC circuit is crucial to analyzing electrical behaviors. When a capacitor is connected to a voltage source, it begins to accumulate electrical charge. This process is called charging, and it progresses according to a specific exponential relationship:

• The voltage across the capacitor, as it charges, increases gradually and does not abrupt.
• The charge on the capacitor at any time is given by the formula: \( q(t) = C \mathcal{E}(1 - e^{-t/RC}) \), where \( \mathcal{E} \) is the EMF, \( C \) is the capacitance, and \( R \) is the resistance.
• Initially, when uncharged, the voltage across the capacitor is zero.

During discharging, if the capacitor is disconnected from the EMF source and connected across just the resistor, it releases its stored energy:

• The charge and voltage decrease exponentially.
• Mathematically, the process can be described by: \( q(t) = Q_0 \, e^{-t/RC} \), where \( Q_0 \) is the initial charge.

Understanding these principles will enable you to solve problems related to capacitor behavior in circuits effectively.

Ohm's Law

Ohm's Law is a fundamental principle in electronics. It defines the relationship between voltage, current, and resistance in an electrical circuit. It is often expressed as:

• \( V = IR \), where \( V \) is the voltage across the resistor, \( I \) is the current flowing through it, and \( R \) is the resistance.
• This law helps in calculating the current, given the voltage and resistance values.

In an RC circuit, when a capacitor is just beginning to charge, the current through the circuit can be easily determined using this law:

• At the initial moment, the capacitor has no voltage across it, so the entire voltage from the EMF source falls across the resistor.
• The calculated current initially is maximum because the resistance is solely affecting the current.

This principle is vital to understanding how currents behave at various stages of capacitor charging and discharging.

Energy Dissipation in Resistors

Resistors in a circuit dissipate energy because they convert electrical energy into heat. The rate at which energy is dissipated in a resistor is called the power in the resistor, which can be calculated using several formulas:

• \( P_R = I^2 R \), where \( I \) is the current through the resistor.
• Also, \( P_R = V^2/R \) or \( P_R = IV \), depending on known values.

In the context of an RC circuit:

• Energy is primarily dissipated initially, when the capacitor is uncharged, and the current is at its maximum.
• Over time, as the capacitor charges, the current decreases, leading to a decrease in the energy dissipation rate.
• Eventually, when the capacitor is fully charged, energy dissipation drops to zero as the current ceases.

This concept explains why the resistor heats up initially and cools down as the circuit reaches a steady state.

Power Calculations in Circuits

Calculating power in an electrical circuit involves determining the rate at which energy is transformed from the power source to perform work.

• The power output from a source can be calculated as follows: \( P_{source} = P_R + P_C \), where \( P_C \) is the power absorbed or stored in the capacitor, and \( P_R \) is the power dissipated in the resistor.
• When a capacitor begins charging, the entire power output initially goes into the resistor since the capacitor does not store energy immediately.
• As charging progresses, more power begins to store in the capacitor and less is dissipated in the resistor.
• At steady state, with a fully charged capacitor, all incoming power charges the capacitor, with none in the resistor.

Understanding these calculations allows you to predict how energy and power change over time in an RC circuit and aids in effective circuit design.