Question

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

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

Step-by-step solution

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 \( R = 5.00 \) \(\Omega\).

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.

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.

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.

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 \[ U = \frac{1}{2} \times 6.00 \times 10^{-6} \times 11.25^2 \approx 3.796 \times 10^{-4} \text{ J} \].

Key concepts

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.