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A capacitor is charged to a potential of \(12.0 \mathrm{~V}\) and is then connected to a voltmeter having an internal resistance of \(3.40 \mathrm{M} \Omega\). After a time of \(4.00 \mathrm{~s}\), the voltmeter reads \(3.0 \mathrm{~V}\). What are (a) the capacitance and (b) the time constant of the circuit?

Short Answer

Expert verified
Capacitance is \(8.47 \times 10^{-7} \text{ F}\). Time constant is \(2.88 \text{ s}\).

Step by step solution

01

Understanding the Problem

We are given a charged capacitor connected to a voltmeter, and over time the voltage changes from \(12.0 \, \text{V}\) to \(3.0 \, \text{V}\) in \(4.00 \, \text{s}\). We need to find the capacitance \(C\) of the capacitor and the time constant \(\tau\).
02

Using the Exponential Decay Formula

The voltage across the capacitor as it discharges through the resistor is given by the formula: \[ V(t) = V_0 e^{-t/\tau} \]where \(V(t)\) is the final voltage, \(V_0\) is the initial voltage, \(t\) is the time, and \(\tau\) is the time constant. Substitute the given values into this formula.
03

Substitute Known Values

Substitute \(V(t) = 3.0 \, \text{V}\), \(V_0 = 12.0 \, \text{V}\), and \(t = 4.00 \, \text{s}\) into the exponential decay formula:\[ 3.0 = 12.0 \times e^{-4.00/\tau} \]
04

Solve for the Time Constant

Rearrange the equation to solve for \(\tau\):- Divide both sides by \(12.0\): \[ \frac{3.0}{12.0} = e^{-4.00/\tau} \]- Simplify it to \(0.25 = e^{-4.00/\tau} \).- Take the natural logarithm on both sides: \[ \ln(0.25) = -\frac{4.00}{\tau} \]- Rearrange to find \(\tau\): \[ \tau = -\frac{4.00}{\ln(0.25)} \]
05

Calculate the Time Constant

Calculate \(\ln(0.25)\) which is approximately \(-1.3863\). Substitute this back into the equation to find \(\tau\): \[ \tau = \frac{4.00}{1.3863} \approx 2.88 \, \text{s} \]
06

Determine the Capacitance

The time constant \(\tau\) is also given by the formula: \( \tau = R \times C \).We know \(R = 3.40 \, \text{M}\Omega = 3.40 \times 10^6 \, \Omega\), and \(\tau\approx 2.88 \, \text{s}\). Solve for \(C\):\[ C = \frac{\tau}{R} = \frac{2.88}{3.40 \times 10^6} \approx 8.47 \times 10^{-7} \, \text{F} \]
07

Final Calculations and Conclusion

Re-calculate if necessary and confirm: - Capacitance \(C\) is approximately \(8.47 \times 10^{-7} \, \text{F}\).- Time constant \(\tau\) is approximately \(2.88 \, \text{s}\).The results match the calculations, confirming the capacitor's capacitance and the time constant.

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Key Concepts

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

Exponential Decay
In a resistor-capacitor circuit, the process of a capacitor discharging over time can be described by an exponential decay. This means that the voltage across the capacitor decreases exponentially as it releases its stored electrical energy. The mathematical formula for this decay is given by:\[ V(t) = V_0 e^{-t/\tau} \]where:
  • \( V(t) \) is the voltage across the capacitor at time \( t \).
  • \( V_0 \) is the initial voltage before discharging starts.
  • \( t \) represents time.
  • \( \tau \) is the time constant of the circuit.
To solve problems involving exponential decay, we substitute known values into this equation. This approach can help us understand how quickly the voltage across a capacitor will drop, crucial for applications where timing is key, such as in timers and delay circuits.
Time Constant
The time constant, represented by the Greek letter \( \tau \), is a critical parameter in a resistor-capacitor (RC) circuit. It tells us how quickly the circuit responds to changes in voltage. Mathematically, it is expressed as the product of resistance \( R \) and capacitance \( C\):\[ \tau = R \times C \]The time constant is important because it determines the rate of charge and discharge of a capacitor. For instance, after one time constant \( \tau \), the voltage across the capacitor will have decreased to about \( 37\% \) (since \( e^{-1} \approx 0.37 \)) of its initial value. Understanding the time constant helps us design circuits to slow down or speed up the response of electronic devices. It's a measure of how quickly the circuit reaches a stable condition, thus influencing dynamics in electronic filters, integrators, and differentiators.
Voltmeter Measurement
Measuring the voltage across a capacitor using a voltmeter gives us instantaneous readings of how much potential energy remains in the capacitor. However, it's crucial to consider the internal resistance of the voltmeter, which can affect measurements in delicate circuits.
  • Voltmeter internal resistance provides a parallel path for discharge.
  • In our exercise, the voltmeter has a resistance of \( 3.40 \mathrm{M} \Omega \).
  • This high resistance ensures it minimally affects the circuit, but it must be considered when calculating the time constant.
To ensure accurate readings, choose voltmeters with a high internal resistance compared to the circuit's resistive elements. This setup minimizes the voltmeter's impact on the circuit operation and allows precise monitoring of voltage changes during discharge. That way, the voltmeter can effectively capture the voltage decay pattern as described by the exponential decay formula.
Resistor-Capacitor Circuit
A resistor-capacitor (RC) circuit is a fundamental configuration in electronics, consisting of a resistor and a capacitor connected in series or parallel to each other. The combination of these two passive components determines how the circuit responds to voltage or current changes over time. In our capacitor discharge example:
  • The capacitor initially stores electrical energy as voltage.
  • When connected to a resistor, the energy is discharged through the resistor, following an exponential decay pattern.
  • The rate at which this discharge occurs is governed by the time constant \( \tau = R \times C \).
Such circuits are widely used in various applications:- As timing elements in clocks.- For smoothing voltage in power supplies.- In audio equipment to filter signals.Understanding the RC circuit allows engineers to design systems that adequately control the timing sequence of electrical processes, enhancing functionality in a wide array of electronic devices.

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Most popular questions from this chapter

A steel wire of length \(L\) and radius \(r_{1}\) has a resistance \(R\). A second steel wire has the same length but a radius \(r_{2}\) and a resistance of \(3 R\). Find the ratio \(r_{1} / r_{2}\)

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A \(6.00 \mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(4500 \Omega\) resistor and a \(500 \mathrm{~V}\) emf source with negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor, (b) the voltage drop across the resistor, (c) the charge on the capacitor, and (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities?

A ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

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