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

Short Answer

Expert verified
Capacitance is approximately \( 8.47 \times 10^{-7} \text{ F} \) and the time constant is 2.88 seconds.

Step by step solution

01

Understand the problem

We are given a circuit with a capacitor initially charged to 12.0 V, which discharges through a voltmeter with an internal resistance of 3.40 MĪ©. The voltmeter reads 3.0 V after 4.00 s. We need to find the capacitance of the capacitor and the time constant of the circuit.
02

Formula for Voltage in RC Circuit

The voltage across a capacitor in a discharging RC circuit is given by the formula:\[ V(t) = V_0 \times e^{-t/RC} \]where \( V_0 \) is the initial voltage, \( V(t) \) is the voltage at time \( t \), \( R \) is the resistance, and \( C \) is the capacitance.
03

Rearrange the formula to solve for RC

We can rearrange the formula to solve for \( RC \):\[ RC = -\frac{t}{\ln(V(t)/V_0)} \]Substitute \( V(t) = 3.0 \) V, \( V_0 = 12.0 \) V, and \( t = 4.00 \) s.
04

Calculate the Time Constant (RC)

Substitute the values into the formula:\[ RC = -\frac{4.00}{\ln(3.0/12.0)} \]Calculate the natural logarithm and solve:\[ \ln(3/12) = \ln(0.25) \approx -1.386 \]\[ RC \approx -\frac{4.00}{-1.386} \approx 2.88 \text{ seconds} \]
05

Use Resistance to Find Capacitance

We know \( RC = 2.88 \) seconds, and the resistance \( R = 3.40 \times 10^6 \) Ī©. Solve for \( C \):\[ C = \frac{2.88}{3.40 \times 10^6} \approx 8.47 \times 10^{-7} \text{ F} \]
06

Summary of Results

The capacitance of the capacitor is approximately \( 8.47 \times 10^{-7} \text{ F} \), and the time constant of the circuit is approximately 2.88 seconds.

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Key Concepts

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

Capacitance
Capacitance is a fundamental property of capacitors that defines their ability to store electrical charge. In simpler terms, capacitance represents how much charge a capacitor can hold for a given voltage across its terminals. In the context of our exercise, the capacitor in question is initially charged to 12.0 V. After being connected to the voltmeter with an internal resistance, the challenge is to determine how much electrical energy it can store given this new configuration.

The relationship between charge (\( Q \)) and voltage (\( V \)) in a capacitor is expressed as \( Q = C \times V \), where \( C \) is the capacitance. Therefore, when we connect this capacitor in an RC circuit, the time it takes to discharge significantly depends on this capacity. By analyzing how the voltage changes over time, we can calculate the capacitance using the changes in voltage and known resistance. This exercise calculated \( C \approx 8.47 \times 10^{-7} \) F, indicating the capacitor's ability to store charge under given conditions.
Time Constant
The time constant (\( \tau \)) in an RC circuit is a key measure of how quickly the circuit responds to changes in voltage. It is essentially the time required for a capacitor to discharge to about 37% of its initial value or to charge up to about 63% of its maximum capacity. This time constant plays a crucial role in determining how long a circuit will take to either charge or discharge.

In mathematical terms, the time constant is the product of resistance (\( R \)) and capacitance (\( C \)) of the circuit: \( \tau = R \cdot C \). In the exercise, this is computed as approximately 2.88 seconds. It tells us that in 2.88 seconds, the voltage across the capacitor in our circuit will have dropped significantly from its initial value. Understanding this concept helps students predict how circuits behave over time and adjust their designs or expectations accordingly.
Voltmeter Resistance
In the context of our circuit, voltmeter resistance refers to the internal resistance present within the voltmeter that is part of the measurement process. When connected across a capacitor, the voltmeter acts as an additional resistor in the circuitā€”here, with a resistance of 3.40 MĪ©.

This internal resistance is crucial because it directly impacts the rate at which the capacitor discharges. A higher resistance will slow down the discharge, leading the voltmeter to show a slower decrease in voltage readings over time. Conversely, a lower resistance would result in a quicker discharge and a more rapid drop in voltage readings. Hence, the voltmeter not only measures the voltage but also influences the circuit's behavior by being part of it. The correct consideration of this resistance is essential for accurate calculations and understanding of the RC circuit's dynamics.

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Most popular questions from this chapter

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?

A resistor with \(R =\) 850 \(\Omega\) is connected to the plates of a charged capacitor with capacitance \(C =\) 4.62 \(\mu\)F. Just before the connection is made, the charge on the capacitor is 6.90 mC. (a) What is the energy initially stored in the capacitor? (b) What is the electrical power dissipated in the resistor just after the connection is made? (c) What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part (a)?

A 2.36-\(\mu\)F capacitor that is initially uncharged is connected in series with a 5.86-\(\Omega\) resistor and an emf source with \(\varepsilon =\) 120 V and negligible internal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value.

A 1.50-\(\mu\)F capacitor is charging through a 12.0-\(\Omega\) resistor using a 10.0-V battery. What will be the current when the capacitor has acquired \\(\frac{1}{4}\\) of its maximum charge? Will it be \\(\frac{1}{4}\\) of the maximum current?

A resistor with \(R_1 =\) 25.0 \(\Omega\) is connected to a battery that has negligible internal resistance and electrical energy is dissipated by \(R_1\) at a rate of 36.0 W. If a second resistor with \(R_2 =\) 15.0 \(\Omega\) is connected in series with \(R_1\), what is the total rate at which electrical energy is dissipated by the two resistors?

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