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A 6.00-\(\mu\)F capacitor that is initially uncharged is connected in series with a 5.00-\(\Omega\) resistor and an emf source with \(\varepsilon =\) 50.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?

Short Answer

Expert verified
3.796 \(\times 10^{-4}\) J.

Step by step solution

01

Understand the Power Dissipation in the Resistor

The problem states that the resistor dissipates energy at a rate of 300 W. The power dissipated by a resistor in a circuit is given by the formula \( P = I^2 R \), where \( I \) is the current through the resistor, and \( R \) is the resistance. Here, \( P = 300 \) W and \( R = 5.00 \) \(\Omega\).
02

Calculate the Current Through the Resistor

Rearrange the power formula to solve for current: \( I = \sqrt{\frac{P}{R}} \). Substituting the given values, \( I = \sqrt{\frac{300}{5}} = \sqrt{60} \approx 7.75 \) A.
03

Determine the Voltage Across the Resistor

The voltage across the resistor can be calculated using Ohm's Law \( V = IR \). Substituting \( I = 7.75 \) A and \( R = 5 \) \(\Omega\), we find \( V = 7.75 \times 5 = 38.75 \) V.
04

Calculate the EMF Source Voltage shared across the Capacitor

The emf source is 50.0 V. At this instant, the voltage across the capacitor \( V_c \) is the remaining voltage after the voltage drop across the resistor: \( V_c = \varepsilon - V = 50 - 38.75 = 11.25 \) V.
05

Calculate the Energy Stored in the Capacitor

The energy stored in a capacitor is given by the formula \( U = \frac{1}{2} C V^2 \). Substituting \( C = 6.00 \times 10^{-6} \) F and \( V_c = 11.25 \) V, we get \[ U = \frac{1}{2} \times 6.00 \times 10^{-6} \times 11.25^2 \approx 3.796 \times 10^{-4} \text{ J} \].

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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 used in electrical circuits to relate voltage, current, and resistance. It states that the current passing through a resistor between two points is directly proportional to the voltage across the two points. This relationship can be expressed by the formula \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.
It's vital to understand Ohm's Law when analyzing circuits because it helps determine how voltage and current distribute in a circuit, especially when components like resistors are involved. In this problem, we used Ohm's Law to determine the voltage drop across a resistor given a current of 7.75 A and a resistance of 5.00 \( \Omega \). This calculation is crucial as it helps identify how much of the total voltage from the emf source is utilized by the resistor versus how much goes across the capacitor.
Power Dissipation
Power dissipation refers to the process by which electrical power is converted into heat in a circuit component, such as a resistor. In resistors, power dissipation is crucial because it affects circuit performance and can impact a device's operating temperature. The formula to calculate power dissipation in a resistor is \( P = I^2 R \), where \( P \) is the power dissipated, \( I \) is the current, and \( R \) is the resistance.
In our example, a resistor dissipates 300 W of power while being part of an electrical circuit with a current of 7.75 A and a resistance of 5.00 \( \Omega \). Calculating power dissipation allows us to measure how much energy is being used at a given moment in a circuit, which is crucial for understanding efficiency and preventing overheating.
Capacitor Discharge
Capacitor discharge is the process by which a charged capacitor releases its stored electrical energy back into the circuit. The speed and duration of this discharge are determined by the resistance in the circuit. In this context, the capacitor is initially charged by the circuit voltage. When the capacitor discharges, it releases energy which can be calculated using \( U = \frac{1}{2} C V^2 \), where \( U \) is the stored energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.
In the problem, once the voltage across the resistor is calculated and subtracted from the emf of the source, the remaining voltage (11.25 V) is the voltage across the capacitor. This allows us to calculate the energy stored as \( 3.796 \times 10^{-4} \) J, which indicates how much energy is available for the capacitor to discharge.
Electrical Circuits
Understanding electrical circuits is essential for resolving problems involving various electronic components such as resistors and capacitors. An electrical circuit is a closed loop that allows current to flow, enabling energy transfer from the source to different components. Components in circuits can be configured in series or parallel arrangements, affecting how voltage and current are distributed.
In this exercise, we dealt with a simple series circuit comprising a resistor and capacitor connected to an emf source. This setup highlights fundamental principles like Ohm's Law and power dissipation, combined with the concept of controlled capacitance through charge and discharge cycles. Analyzing these components' interactions provides insights into the stored energy and power dynamics in electrical circuits, which is crucial for designing efficient and safe electronic devices.

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Most popular questions from this chapter

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 W, 900 W, 1200 W, and 1500 W. You start with the hair dryer on the 600-W setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-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 kWh, how much does it cost per hour to operate the dryer?

A 4.60-\(\mu\)F capacitor that is initially uncharged is connected in series with a 7.50-k\(\Omega\) resistor and an emf source with \(\varepsilon =\) 245 V and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)?

A circuit consists of a series combination of 6.00-k\(\Omega\) and 5.00-k\(\Omega\) resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00-k\(\Omega\) resistor using a voltmeter having an internal resistance of 10.0 k\(\Omega\). (a) What potential difference does the voltmeter measure across the 5.00-k\(\Omega\) resistor? (b) What is the \(true\) potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?

A galvanometer having a resistance of 25.0 \(\Omega\) has a 1.00-\(\Omega\) shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a circuit consisting of a 15.0-\(\Omega\) resistor connected across the terminals of a 25.0-V battery having no appreciable internal resistance. (a) What current does the ammeter measure? (b) What should be the \(true\) current in the circuit (that is, the current without the ammeter present)? (c) By what percentage is the ammeter reading in error from the \(true\) current?

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