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A \(3.0-\mathrm{M} \Omega\) resistor is connected in series with an initially uncharged \(0.28-\mu \mathrm{F}\) capacitor. This arrangement is then connected across four \(1.5-\mathrm{V}\) batteries (also in series). (a) What is the maximum current in the circuit and when does it occur? (b) What percentage of the maximum current is in the circuit after 4.0 s? (c) What is the maximum charge on the capacitor and when does it occur? (d) What percentage of the maximum charge is on the capacitor after 4.0 s? (e) How much energy is stored in the capacitor after one time constant has elapsed?

Short Answer

Expert verified
(a) Maximum current is \(2.0 \times 10^{-6}\) A at \(t = 0\). (b) 0.5\% of maximum current after 4.0 s. (c) Maximum charge is \(1.68 \times 10^{-6}\) C at full charge. (d) Approximately 100\% charge after 4.0 s. (e) Energy stored is \(2.01 \times 10^{-6}\) J after one time constant.

Step by step solution

01

Maximum Current in the Circuit

The maximum current in a charging RC circuit occurs when the capacitor is uncharged and is calculated using Ohm's Law: \( I_{max} = \frac{V}{R} \). Here, \( V = 4 \times 1.5 \, V = 6.0 \, V \) and \( R = 3.0 \, M\Omega = 3.0 \times 10^6 \, \Omega \). Thus, the maximum current \( I_{max} = \frac{6.0}{3.0 \times 10^6} = 2.0 \times 10^{-6} \, A \). It occurs at \( t = 0 \) s, as that's when the capacitor is not yet charged at all.
02

Current After 4.0 Seconds

The current at any time \( t \) in an RC circuit is given by: \( I(t) = I_{max} e^{-t/RC} \). First, calculate the time constant \( \tau = RC = (3.0 \times 10^6)(0.28 \times 10^{-6}) = 0.84 \, s \). Plug in the values: \( I(4) = 2.0 \times 10^{-6} e^{-4.0/0.84} \approx 1.0 \times 10^{-8} \, A \). Thus, the percentage of the maximum current after 4.0 s is \( \frac{1.0 \times 10^{-8}}{2.0 \times 10^{-6}} \times 100 \approx 0.5\% \).
03

Maximum Charge on the Capacitor

The maximum charge \( Q_{max} \) on the capacitor is given by \( Q_{max} = C \times V \). With \( C = 0.28 \times 10^{-6} \, F \) and \( V = 6.0 \, V \), the charge is \( Q_{max} = 0.28 \times 10^{-6} \times 6.0 = 1.68 \times 10^{-6} \, C \). This occurs as \( t \to \infty \).
04

Charge After 4.0 Seconds

The charge on the capacitor after time \( t \) is given by \( Q(t) = Q_{max} (1 - e^{-t/RC}) \). Using values from previous steps: \( Q(4) = 1.68 \times 10^{-6} (1 - e^{-4.0/0.84}) \approx 1.68 \times 10^{-6} \, C \). This charge is approximately 100\% of the maximum charge, indicating full charge of the capacitor after 4.0 s.
05

Energy Stored After One Time Constant

The energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \). After one time constant (\( \tau = 0.84 \, s \)), the capacitor is about 63.2\% charged. Thus, the voltage across is \( 0.632 \times 6.0 \approx 3.792 \, V \). Calculating energy: \( E \approx \frac{1}{2} (0.28 \times 10^{-6}) (3.792)^2 \approx 2.01 \times 10^{-6} \, J \).

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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 an essential principle in electronics, relating voltage (V), current (I), and resistance (R) in an electric circuit. According to this law, the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance.
The formula for Ohm's Law is given by:
  • Current (\( I \)) = Voltage (\( V \)) / Resistance (\( R \))
In the context of an RC circuit—comprising a resistor and a capacitor in series—Ohm's Law helps determine the maximum current in the circuit right when the process starts.
At the initial moment when the capacitor is still uncharged, Ohm’s Law tells us the maximum current is achieved, calculated simply as the total voltage divided by the resistance.
In our exercise, with a total voltage of 6.0 V and a resistance of 3.0 MegaOhms, the initial maximum current is computed as:
  • \( I_{max} = \frac{6.0 \,V}{3.0 \times 10^6 \, \Omega} = 2.0 \times 10^{-6} \, A \)
Time Constant
The time constant in an RC circuit is a critical parameter that determines how quickly the circuit responds to changes, such as charging and discharging of a capacitor.
It is represented by the Greek letter \( \tau \) (tau) and mathematically, it is the product of the resistance (R) and the capacitance (C).
  • Time Constant (\( \tau \)) = Resistance (\( R \)) × Capacitance (\( C \))
For our circuit with a resistance of 3.0 MegaOhms and a capacitance of 0.28 microFarads, the time constant is:
  • \( \tau = (3.0 \times 10^6) \times (0.28 \times 10^{-6}) = 0.84 \, s \)
The time constant indicates the time it takes for the charge on the capacitor to reach about 63.2% of its maximum value during charging.
This concept is crucial, providing a time frame for when major changes occur in the circuit’s behavior.
Capacitor Charge
Capacitors are components that store electric charge and energy in an electric field. The amount of charge a capacitor can hold is determined by its capacitance and the voltage applied across it.
The charge on a capacitor at any point in time during charging or discharging can be modeled and predicted using exponential functions because, in RC circuits, these processes are not linear.
The maximum charge (\( Q_{max} \)) a capacitor can hold, when fully charged, can be calculated by multiplying the capacitance (\( C \)) by the voltage (\( V \)) across it.
  • Maximum Charge (\( Q_{max} \)) = Capacitance (\( C \)) × Voltage (\( V \))
    • \( Q_{max} = 0.28 \times 10^{-6} \times 6.0 = 1.68 \times 10^{-6} \, C \)
This maximum charge occurs as time approaches infinity, illustrating that perfectly complete charging is theoretical and takes an indefinite period.
Energy Storage in Capacitors
Capacitors store energy in an electric field, and the energy stored within can be calculated using the formula for the energy of a charged capacitor. The formula is given as:
  • \( E = \frac{1}{2} C V^2 \)
Where:
  • \( E \) is the energy stored
  • \( C \) is the capacitance
  • \( V \) is the voltage across the capacitor
After a certain period, specifically one time constant (\( \tau \)), the capacitor reaches about 63.2% of its total charge, allowing us to also determine the energy stored at this point.
For our specific circuit, using a voltage reduced to 63.2% of 6.0 V, the energy stored is calculated to demonstrate how capacitors accumulate energy over time, illustrating the gradual nature of charging:
  • \( E \approx \frac{1}{2} \times 0.28 \times 10^{-6} \times (3.792)^2 \approx 2.01 \times 10^{-6} \, J \)

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