Question

A \(2.00-\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(6.00-\mathrm{k} \Omega\) resistor and an emf source with \(\mathcal{E}=90.0 \mathrm{~V}\) and negligible internal resistance. The circuit is completed at \(t=0 .\) (a) Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor? (b) At what value of \(t\) is the rate at which electrical energy is being dissipated in the resistor equal to the rate at which electrical energy is being stored in the capacitor? (c) At the time calculated in part (b), what is the rate at which electrical energy is being dissipated in the resistor?

Short answer

(a) 1.35 W; (b) At \(t = 8.316 \times 10^{-3} \) s; (c) 1.35 W.

Step-by-step solution

Understand the circuit and given information

We have a series circuit containing a capacitor and a resistor connected to an emf (electromotive force) source. The capacitor has a capacitance of \( C = 2.00 \, \mu \mathrm{F} \) (microfarads), the resistor has a resistance of \( R = 6.00 \, \mathrm{k} \Omega \), and the battery's emf is \( \mathcal{E} = 90.0 \, \mathrm{V} \). The internal resistance of the emf source is negligible.

Calculate the initial rate of energy dissipation in the resistor

Right after the switch is closed at \( t = 0 \), the capacitor is uncharged, meaning the entire voltage \( \mathcal{E} = 90.0 \, \mathrm{V} \) drops across the resistor.The rate of energy dissipation in a resistor is given by the power formula \( P = \frac{V^2}{R} \).Substitute: \( P = \frac{(90.0 \, \mathrm{V})^2}{6.00 \, \mathrm{k} \Omega} = \frac{8100}{6000} = 1.35 \, \mathrm{W} \).

Determine when the power dissipation equals energy storage

The energy dissipated in the resistor is given by \( P_r(t) = I(t)^2 R \), where \( I(t) \) is the current. The energy stored in the capacitor is \( P_c(t) = \frac{d}{dt} \frac{1}{2}CV(t)^2 = CV(t)\frac{dV(t)}{dt} \), where \( V(t) \) is the voltage across the capacitor.When these two powers are equal, solve \( I(t)^2 R = CV(t)\frac{dV(t)}{dt} \). Solving gives time constant \( t = RC\ln(2) \).

Calculate the time when the powers are equal

Plug in given values: \( R = 6.00 \, \mathrm{k} \Omega = 6000 \, \Omega \) and \( C = 2.00 \, \mu \mathrm{F} = 2.00 \times 10^{-6} \, \mathrm{F} \).Use \( t = RC \ln(2) = 6000 \times 2.00 \times 10^{-6} \times \ln(2) \).This simplifies to: \( t = 6000 \times 2.00 \times 10^{-6} \times 0.693 = 8.316 \times 10^{-3} \, \mathrm{s} \).

Calculate the power dissipation at the calculated time

At \( t = 8.316 \times 10^{-3} \, \mathrm{s} \), use \( I(t) = \frac{\mathcal{E}}{R} \cdot e^{-t/RC} \).Using \( I(t) = \frac{90.0}{6000} \cdot e^{-8.316 \times 10^{-3}/(6000\times2.00\times10^{-6})} \).Solve for \( I(t) \) to find the power \( P_r(t) = I(t)^2 R \). Calculate \( I(t) = 0.015\,\mathrm{A} \) leading to \( P = (0.015)^2 \times 6000 = 1.35 \, \mathrm{W} \).

Key concepts

Capacitor Charging

When a capacitor charges in a circuit, it accumulates electrical energy over time. At the start, when a switch is turned on, the voltage from the battery causes a current to begin flowing through the circuit. This initial current is at its maximum because the capacitor is uncharged and offers no resistance to the current flow.
The voltage across the capacitor increases as it charges, and thus, less current can flow through the resistor using the formula \( V(t) = \mathcal{E}(1 - e^{-t/RC}) \). Here, \( V(t) \) is the voltage across the capacitor at time \( t \), and \( e \) represents the exponential constant. The equation tells us that the charging curve of a capacitor is exponential, meaning it charges quickly at first, and then the rate of charging slows down markedly.

• Initial voltage across the capacitor: 0 V (unstarched)
• Eventually reaches: 90 V (the battery voltage)

Energy Dissipation

In electric circuits, energy dissipation usually occurs in the form of heat. This is because resistors convert electrical energy into thermal energy. At the moment the circuit is completed, energy is dissipated in the resistor at a peak rate. This is governed by the equation \( P = \frac{V^2}{R} \), where \( P \) is the power dissipated, \( V \) is the voltage across the resistor, and \( R \) is the resistance.
Since the voltage at the start is completely across the resistor, the initial power dissipation is its highest. As time goes on and the capacitor charges, the voltage across the resistor drops, leading to a decrease in the power dissipated as heat. Hence, the dissipation rate decreases gradually.

• Initially High Power: Higher voltage across the resistor
• Decreasing over Time: As the capacitor takes over part of the voltage

Time Constant

The time constant, frequently denoted as \( \tau \), is a fundamental parameter in RC circuits. This value is found by multiplying the resistance \( R \) by the capacitance \( C \) of the circuit: \( \tau = RC \). The time constant tells us how fast the capacitor charges or discharges.
After one time constant \( \tau \), the voltage across the capacitor reaches about 63.2% of its maximum value, and the remaining charge happens at a diminishing rate. The time constant is crucial because, after about 5 time constants, the capacitor is almost completely charged at about 99%.

• One Time Constant: Reaches 63.2% charge
• Five Time Constants: Nearly fully charged

In our exercise, \( \tau \) is calculated as follows: \( \tau = 6.00 \times 10^3 \,\Omega \times 2.00 \times 10^{-6} \,\mathrm{F} = 1.2 \times 10^{-2} \,\mathrm{s} \).

Electrical Energy Conversion

During the charging process of the capacitor, electrical energy from the circuit's power source is converted and stored as electrostatic potential energy in the capacitor. The stored energy can later be released back to the circuit.
The energy stored in a capacitor can be calculated with the formula \( U = \frac{1}{2}CV^2 \). Here, \( U \) represents the stored energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. As the capacitor charges, it does so by converting some energy from the resistor to stored energy.
Important points:

• Initially, most energy goes to resistor dissipation instead of storage
• As the capacitor charges, conversion into stored energy increases