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Chapter 27: Problem 58

In an \(R C\) series circuit, emf \(\&=12.0 \mathrm{~V}\), resistance \(R=\) \(1.40 \mathrm{M} \Omega\), and capacitance \(C=1.80 \mu \mathrm{F}\). (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to \(16.0 \mu \mathrm{C}\) ?

Short Answer

Expert verified
(a) 2.52 s, (b) 21.6 渭C, (c) 3.40 s.

Step by step solution

01

Calculating the Time Constant

The time constant \( \tau \) of an \( R C \) circuit is given by \( \tau = R \cdot C \). Substitute the given values: \( R = 1.40 \times 10^6 \) ohms and \( C = 1.80 \times 10^{-6} \) farads. Therefore, \( \tau = 1.40 \times 10^6 \times 1.80 \times 10^{-6} = 2.52 \) seconds.
02

Finding the Maximum Charge on the Capacitor

The maximum charge \( Q_{max} \) on a capacitor is given by \( Q_{max} = C \cdot \text{emf} \). Substituting \( C = 1.80 \times 10^{-6} \) farads and \( \text{emf} = 12.0 \) volts, we find \( Q_{max} = 1.80 \times 10^{-6} \times 12.0 = 21.6 \mu \text{C} \).
03

Calculating the Time to Reach a Specific Charge

The charge \( Q \) on a capacitor as it charges over time \( t \) is given by the equation \( Q = Q_{max}(1 - e^{-t/\tau}) \). We need to find the time \( t \) when \( Q = 16.0 \mu \text{C} \). Use the equation: \[ 16.0 \mu \text{C} = 21.6 \mu \text{C} (1 - e^{-t/2.52}) \]Rearrange to solve for \( t \): \[ 1 - \frac{16.0}{21.6} = e^{-t/2.52} \] \[ e^{-t/2.52} = 0.259 \]Taking the natural logarithm on both sides: \[ -t/2.52 = \ln(0.259) \approx -1.35 \] Thus, \[ t = 2.52 \times 1.35 = 3.40 \text{ seconds} \].

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Key Concepts

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

Time Constant
In an RC circuit, the time constant, denoted by \( \tau \), is a measure of time it takes for a capacitor to charge up to about 63% of its full capacity or discharge to about 37% of its full charge. It is a crucial parameter because it dictates how quickly or slowly the capacitor responds to the electric current.
For an RC circuit, the time constant is calculated using the formula \( \tau = R \cdot C \), where \( R \) is resistance in ohms and \( C \) is capacitance in farads.
From the problem, with a resistance of \( 1.40 \times 10^6 \) ohms and a capacitance of \( 1.80 \times 10^{-6} \) farads, we find that \( \tau = 2.52 \) seconds.
  • This means it takes about 2.52 seconds for the capacitor to reach 63% of its maximum charge.
  • The time constant helps us predict the behavior of the charging and discharging process of capacitors.
Maximum Charge
The maximum charge that a capacitor can hold when fully charged, referred to as \( Q_{max} \), occurs when the capacitor voltage equals the voltage of the source.
It is calculated by the formula \( Q_{max} = C \cdot \text{emf} \), where \( C \) is the capacitance and \( \text{emf} \) is the electromotive force in volts.
  • In this problem, with \( C = 1.80 \times 10^{-6} \) farads and \( \text{emf} = 12.0 \) volts, the maximum charge \( Q_{max} \) is found to be \( 21.6 \mu \text{C} \).
Understanding \( Q_{max} \) is essential as it tells us the full capacity of the capacitor, which is crucial for applications where specific charge levels must not be exceeded or need to be reliably achieved.
Capacitor Charging Time
Understanding the time taken to charge a capacitor to a specific charge level involves the relationship with the time constant.
The charge \( Q \) on a capacitor at any time \( t \) can be calculated with the formula: \[ Q = Q_{max}(1 - e^{-t/\tau}) \]
  • Here, \( Q_{max} \) is the maximum charge the capacitor can hold, and \( \tau \) is the time constant.
  • In this scenario, to determine the time needed for the capacitor to reach \( 16.0 \mu \text{C} \), we rearrange the formula and solve: \[ t = \tau \cdot \ln \left(\frac{Q_{max}}{Q_{max} - Q}\right) \]
By substituting values, we find \( t \approx 3.40 \text{ seconds} \).
Understanding this can help in designing circuits where precise timing to charge a capacitor is necessary, like in timing circuits or signal processing.

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Most popular questions from this chapter

A \(15,0 \mathrm{k} \Omega\) resistor and a capacitor are connected in series, and then a \(12.0 \mathrm{~V}\) potential difference is suddenly applied across them. The potential difference across the capacitor rises to \(5.00 \mathrm{~V}\) in \(1.30 \mu \mathrm{s}\) (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.

You are given a number of \(10 \Omega\) resistors, each capable of dissipating only \(1.0 \mathrm{~W}\) without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a \(10 \Omega\) resistance that is capable of dissipating at least \(5.0 \mathrm{~W} ?\)

The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other) \(2.0 \mu \mathrm{F}\) capacitor drops to one-fourth its initial value in \(2.0 \mathrm{~s}\). What is the equivalent resistance between the capacitor plates?

Side flash. Figure \(27-38\) indicates one reason no one should stand under a tree during a lightning storm. If lightning comes down the side of the tree, a portion can jump over to the person, especially if the current on the tree reaches a dry region on the bark and thereafter must travel through air to reach the ground. In the figure, part of the lightning jumps through distance \(d\) in air and then travels through the person (who has negligible resistance relative to that of air because of the highly conducting salty fluids within the body). The rest of the current travels through air alongside the tree, for a distance \(h\). If \(d / h=0.400\) and the total current is \(I=5000 \mathrm{~A}\), what is the current through the person?

A \(3.00 \mathrm{M} \Omega\) resistor and 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?
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