/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Displays two circuits with a cha... [FREE SOLUTION] | 91Ó°ÊÓ

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Displays two circuits with a charged capacitor that is to be discharged through a resistor when a switch is closed. In Fig. 27-38a, \(R_{1}=20.0 \Omega\) and \(C_{1}=5.00 \mu \mathrm{F}\). In Fig. \(27-38 b, R_{2}=10.0 \Omega\) and \(C_{2}=8.00 \mu \mathrm{F}\). The ratio of the initial charges on the two capacitors is \(q_{02} / q_{01}=1.75\). At time \(t=0\), both switches are closed. At what time \(t\) do the two capacitors have the same charge?

Short Answer

Expert verified
The two capacitors have the same charge at approximately 0.224 seconds.

Step by step solution

01

Understanding Capacitor Discharge

When a capacitor discharges through a resistor, the charge on the capacitor decreases over time according to the formula: \[ q(t) = q_0 e^{-t/RC} \] where \( q_0 \) is the initial charge, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is the time. The charge decreases exponentially with a time constant \( \tau = RC \).
02

Write the Charge Equations for Both Circuits

For the first circuit, the charge equation is: \[ q_1(t) = q_{01} e^{-t/(R_1 C_1)} \] For the second circuit, the charge equation is: \[ q_2(t) = q_{02} e^{-t/(R_2 C_2)} \] Given the initial charge ratio, \( q_{02} / q_{01} = 1.75 \).
03

Equate the Charges at Time t

To find when the capacitors have the same charge, set the two charge equations equal to each other:\[ q_{01} e^{-t/(R_1 C_1)} = q_{02} e^{-t/(R_2 C_2)} \] Substitute \( q_{02} = 1.75 q_{01} \) from the given ratio to simplify the equation.
04

Solve for Time t

With substitution, the equation becomes: \[ e^{t/(R_2 C_2)} = 1.75 e^{t/(R_1 C_1)} \] Taking natural logarithms on both sides gives:\[ \frac{t}{R_2 C_2} = \ln(1.75) + \frac{t}{R_1 C_1} \]Rearrange to solve for \( t \):\[ t \left( \frac{1}{R_2 C_2} - \frac{1}{R_1 C_1} \right) = \ln(1.75) \]Substitute \( R_1 = 20 \Omega \), \( C_1 = 5 \times 10^{-6} \mathrm{F} \), \( R_2 = 10 \Omega \), \( C_2 = 8 \times 10^{-6} \mathrm{F} \), and \( \ln(1.75) \approx 0.5596 \) to find \( t \).
05

Calculate Time t

Substitute the known values into the rearranged formula:\[ t \left( \frac{1}{10 \times 8 \times 10^{-6}} - \frac{1}{20 \times 5 \times 10^{-6}} \right) = 0.5596 \]Calculate the exponents:\[ t \left( \frac{1}{80 \times 10^{-6}} - \frac{1}{100 \times 10^{-6}} \right) = 0.5596 \]\[ t \left( 12.5 - 10 \right) = 0.5596 \]\[ 2.5t = 0.5596 \]Finally, solve for \( t \):\[ t = \frac{0.5596}{2.5} = 0.22384 \] seconds.

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Key Concepts

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

Exponential Decay
Exponential decay is a fundamental concept in physics and mathematics. It describes how quantities decrease rapidly at first, then more slowly over time. In the context of capacitors, when one discharges through a resistor, the charge (or energy stored) on the capacitor decreases exponentially.
This process is mathematically represented by the equation:
  • \( q(t) = q_0 e^{-t/RC} \)
Here, \( q(t) \) represents the charge on the capacitor at time \( t \), \( q_0 \) is the initial charge, \( R \) is the resistance, and \( C \) is the capacitance. The variable \( t \) represents the elapsed time. The exponential nature of the function, denoted by \( e^{-t/RC} \), means that the rate at which the charge decreases is proportional to its current value.
This formula shows that as time increases, the term \( e^{-t/RC} \) – an exponential function – rapidly reduces to zero, causing the capacitor charge to drop correspondingly.
Time Constant
The time constant, represented by the Greek letter \( \tau \) (tau), is a key parameter in analyzing the dynamics of both capacitors and resistors in a circuit. It is defined as the product of resistance \( R \) and capacitance \( C \), given by the formula:
  • \( \tau = RC \)
The time constant provides valuable insight into how quickly a capacitor charges or discharges.
  • A small value of \( \tau \) indicates rapid charging or discharging.
  • A larger \( \tau \) means the process is slower, indicating more time is needed for significant charge transfer.
Generally, a capacitor will reach about 63.2% of its total discharge after one time constant (\( \tau \)), and over 99% after approximately five time constants. Understanding the time constant helps predict how quickly a capacitor can respond to changes in a circuit, which is essential for designing time-sensitive applications.
Circuit Analysis
Capacitor discharge analysis is a critical aspect of circuit analysis. By understanding how components interact, you can predict the behavior of a circuit over time. In exercises like the one at hand, analysis begins by identifying the given elements — resistors, capacitors and their initial conditions.
The primary goal of circuit analysis during capacitor discharge is to determine how the charge changes with time. One does this by applying equations that incorporate both the exponential decay pattern and the time constant \( \tau \). For the scenario discussed, two circuits are monitored to find the time \( t \) at which they both have the same charge, necessitating solving for \( t \) where:
  • \( q_{01} e^{-t/(R_1 C_1)} = q_{02} e^{-t/(R_2 C_2)} \)
Prior to solving, circuit analysis demands substituting known values like resistances and capacitances, and handling exponential/logarithmic expressions. These steps enable you to arrive at a specific time \( t \), elucidating when both capacitors achieve an equal charge. This insight forms the basis for more complex analysis and helps foster a deep understanding of circuit behavior.

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Most popular questions from this chapter

Shows the circuit of a flashing lamp, like those attached to barrels at highway construction sites. The fluorescent lamp \(L\) (of negligible capacitance) is connected in parallel across the capacitor \(C\) of an \(R C\) circuit. There is a current through the lamp only when the potential difference across it reaches the breakdown voltage \(V_{L}\); then the capacitor discharges completely through the lamp and the lamp flashes briefly. For a lamp with breakdown voltage \(V_{\mathrm{L}}=75.0 \mathrm{~V}\), wired to a \(95.0 \mathrm{~V}\) ideal battery and a \(0.150 \mu \mathrm{F}\) capacitor, what resistance \(R\) is needed for two flashes per second?

The ideal batteries have emfs \(\&_{1}=200 \mathrm{~V}\) and \(8_{2}=50 \mathrm{~V}\) and the resistances are \(R_{1}=3.0 \Omega\) and \(R_{2}=2.0 \Omega\). If the potential at \(P\) is \(100 \mathrm{~V}\), what is it at \(Q\) ?

A simple ohmmeter is made by connecting a \(9.00 \mathrm{~V}\) flashlight battery in series with a resistance \(R\) and an ammeter that reads from 0 to \(1.00 \mathrm{~mA}\), as shown in Fig. 27-33. Resistance \(R\) is adjusted so that when the clip leads are shorted together, the meter deflects to its full- scale value of \(1.00 \mathrm{~mA}\). What external resistance across the leads results in a deflection of (a) \(10.0 \%\), (b) \(50.0 \%\), and (c) \(90.0 \%\) of full scale? (d) If the ammeter has a resistance of \(20.0 \Omega\) and the internal resistance of the battery is negligible, what is the value of \(R\) ?

A capacitor with initial charge \(q_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first \(25 \%\) of its charge and (b) \(50 \%\) of its charge?

A standard flashlight battery can deliver about \(2.0 \mathrm{~W} \cdot \mathrm{h}\) of energy before it runs down. (a) If a battery costs US\$ \(0.85\), what is the cost of operating a \(100 \mathrm{~W}\) lamp for \(8.0 \mathrm{~h}\) using batteries? (b) What is the cost if energy is provided at the rate of US\$0.06 per kilowatt-hour?

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