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A \(3.00 \mathrm{M} \Omega\) resistor and \(\mathrm{a} 1.00 \mu \mathrm{F}\) capacitor are connected in series with an ideal battery of emf \(\mathscr{E}=4.00 \mathrm{~V}\). At \(1.00 \mathrm{~s}\) after the connection is made, what is the rate at which (a) the charge of the capacitor is increasing, (b) energy is being stored in the capacitor, (c) thermal energy is appearing in the resistor, and (d) energy is being delivered by the battery?

Short Answer

Expert verified
(a) Charge increase rate: 943.62 nA, (b) Energy stored rate: 1.16 µW, (c) Thermal energy rate: 2.67 µW, (d) Energy delivered rate: 3.83 µW.

Step by step solution

01

Calculate the charge on the capacitor at t = 1s

The charge on the capacitor as a function of time when charging is given by\[q(t) = C \mathscr{E} \left(1 - e^{-\frac{t}{RC}}\right),\]where \( C = 1.00\, \mu F \), \( R = 3.00\, M\Omega \), \( \mathscr{E} = 4.00\, V \), and \( t = 1.00\, s \). First, calculate the time constant \( \tau = RC = 3 \times 10^6 \times 1 \times 10^{-6} = 3 \) seconds. Now,\[q(1) = 1 \times 10^{-6} \times 4 \left(1 - e^{-\frac{1}{3}}\right) \approx 1.23 \times 10^{-6}\, C.\]
02

Calculate the rate of change of charge (a)

The rate of change of charge (current) \( i(t) \) is given by the derivative\[i(t) = \frac{dq(t)}{dt} = \frac{\mathscr{E}}{R} e^{-t/RC}\]\[i(1) = \frac{4}{3 \times 10^6} e^{-1/3} \approx 943.62 \times 10^{-9} A \].So, its magnitude is \( 943.62 \, nA \).
03

Calculate the rate energy is stored in the capacitor (b)

The rate at which energy is stored in the capacitor is given by\[ P_{\text{cap}} = q(t) \cdot \frac{dq(t)/dt}{C}\]Substituting from previous calculations:\[P_{\text{cap}} = (1.23 \times 10^{-6}) \cdot (943.62 \times 10^{-9}) / (1.00 \times 10^{-6}) \approx 1.16 \times 10^{-6} W.\]
04

Calculate the rate thermal energy is appearing in the resistor (c)

The power dissipated in the resistor is the thermal energy rate given by: \[ P_{\text{resistor}} = i(t)^2 \cdot R = (943.62 \times 10^{-9})^2 \cdot 3 \times 10^6 \approx 2.67 \times 10^{-6} \text{ W}.\]
05

Calculate the rate energy is delivered by the battery (d)

The total energy delivered by the battery rate is the sum of energy stored in the capacitor and that dissipated by the resistor:\[ P_{\text{battery}} = P_{\text{cap}} + P_{\text{resistor}} = 1.16 \times 10^{-6} + 2.67 \times 10^{-6} = 3.83 \times 10^{-6} \text{ W}.\]

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Key Concepts

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

Capacitor Charging
Understanding how a capacitor charges is essential when dealing with RC circuits. A capacitor is an electrical component that stores energy in an electric field, created by a pair of conductors separated by an insulator. When connected in a circuit with a voltage source, like a battery, it starts charging.The charging process can be described by an exponential function:\[ q(t) = C \mathscr{E} \left(1 - e^{-t/RC}\right) \]where:
  • \( q(t) \) is the charge on the capacitor at time \( t \).
  • \( C \) is the capacitance, measured in farads.
  • \( \mathscr{E} \) is the electromotive force (EMF) of the battery, measured in volts.
  • \( R \) is the resistance, measured in ohms.
  • \( t \) is the time, measured in seconds.
As time progresses, the capacitor approaches its maximum charge, \( C\mathscr{E} \), asymptotically. The time constant, \( \tau = RC \), determines how quickly the capacitor charges. Generally, it takes about 5\( \tau \) for a capacitor to charge up to over 99% of its final value.
Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics and is particularly useful when analyzing RC circuits. It states that the voltage across a resistor is directly proportional to the current flowing through it, with resistance being the factor of proportionality.Mathematically, it is expressed as:\[ V = IR \]where:
  • \( V \) is the voltage drop across the resistor.
  • \( I \) is the current flowing through the resistor.
  • \( R \) is the resistance.
In the context of RC circuits, Ohm's Law helps in determining the current at any point while the capacitor is charging. For example, the initial current when the capacitor is fully discharged is given by \( I = \frac{\mathscr{E}}{R} \). As the capacitor charges, the current decreases exponentially, which can be described by the equation:\[ i(t) = \frac{\mathscr{E}}{R} e^{-t/RC} \]This illustrates how the current draw from the battery decreases as the capacitor reaches its maximum charge.
Energy Dissipation in Resistors
When current flows through a resistor, energy is dissipated as heat due to the resistor's opposition to the flow of electric charge. This process is known as energy dissipation in resistors, and it is a central concern when working with circuits, particularly because it affects the efficiency and thermal management of the system.The power dissipated in a resistor as heat can be calculated using:\[ P = I^2R \]where:
  • \( P \) is the power dissipated.
  • \( I \) is the current flowing through the resistor.
  • \( R \) is the resistance.
In an RC circuit, as the capacitor charges, the instantaneous power dissipated by the resistor can be found by combining this formula with the current expression derived from Ohm's Law:\[ P(t) = \left( \frac{\mathscr{E}}{R} e^{-t/RC} \right)^2 R \]This expression reveals that the power dissipated, or thermal energy rate, decreases over time as the capacitor reaches its steady-state value. It emphasizes the dynamic interplay between charge storage in capacitors and energy dissipation in resistive components.

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Most popular questions from this chapter

(a) In electron-volts, how much work does an ideal battery with a \(12.0 \mathrm{~V}\) emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If \(3.40 \times 10^{18}\) electrons pass through each second, what is the power of the battery in watts?

A \(5.0 \mathrm{~A}\) current is set up in a circuit for \(6.0 \mathrm{~min}\) by a rechargeable battery with a \(6.0 \mathrm{~V}\) emf. By how much is the chemical energy of the battery reduced?

The starting motor of a car is turning too slowly, and the mechanic has to decide whether to replace the motor, the cable, or the battery. The car's manual says that the \(12 \mathrm{~V}\) battery should have no more than \(0.020 \Omega\) internal resistance, the motor no more than \(0.200 \Omega\) resistance, and the cable no more than \(0.040 \Omega\) resistance. The mechanic turns on the motor and measures \(11.4 \mathrm{~V}\) across the battery, \(3.0 \mathrm{~V}\) across the cable, and a current of \(50 \mathrm{~A}\). Which part is defective?

A solar cell generates a potential difference of \(0.10 \mathrm{~V}\) when a \(500 \Omega\) resistor is connected across it, and a potential difference of \(0.15 \mathrm{~V}\) when a \(1000 \Omega\) resistor is substituted. What are the (a) internal resistance and (b) emf of the solar cell? (c) The area of the cell is \(5.0 \mathrm{~cm}^{2},\) and the rate per unit area at which it receives energy from light is \(2.0 \mathrm{~mW} / \mathrm{cm}^{2}\). What is the efficiency of the cell for converting light energy to thermal energy in the \(1000 \Omega\) external resistor?

A car battery with a \(12 \mathrm{~V} \mathrm{emf}\) and an internal resistance of \(0.040 \Omega\) is being charged with a current of \(50 \mathrm{~A}\). What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{\underline{r}}\) of energy dissipation inside the battery, and (c) the rate \(P_{\mathrm{emf}}\) of energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) \(V\) and (e) \(P_{r} ?\)

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