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Chapter 27: Problem 61

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.

Short Answer

Expert verified
(a) The time constant \(\tau\) is \(3.563 \times 10^{-6}\) seconds. (b) The capacitance \(C\) is approximately \(237.5\, \text{pF}\).

Step by step solution

01

Understanding the Time Constant Formula

The time constant \(\tau\) in an RC circuit is given by \(\tau = RC\), where \(R\) is the resistance, and \(C\) is the capacitance. We need to find \(C\) using the given values.
02

Applying the Capacitor Charging Formula

The formula for the voltage \(V(t)\) across the capacitor as a function of time \(t\) during charging is \(V(t) = V_0(1 - e^{-t/\tau})\). Here, \(V_0\) is the initial potential difference applied, which is \(12.0\, \text{V}\). We need to find \(\tau\) that satisfies the condition \(V(1.30 \mu s) = 5.00 \text{V}\).
03

Plugging in Known Values

Set up the equation \(5.00 = 12.0 (1 - e^{-1.30 \times 10^{-6}/\tau})\). Solve for \(\tau\) using logarithmic manipulation.
04

Solving for the Time Constant \(\tau\)

Rearrange and solve the equation: \(e^{-1.30 \times 10^{-6}/\tau} = \frac{12.0 - 5.00}{12.0}\). Taking natural logarithms of both sides gives \(-\frac{1.30 \times 10^{-6}}{\tau} = \ln(\frac{7}{12})\). Solve for \(\tau\): \(\tau = -\frac{1.30 \times 10^{-6}}{\ln(\frac{7}{12})}\).
05

Calculate \(\tau\)

Calculate \(\ln\left(\frac{7}{12}\right)\), then divide \(-1.30 \times 10^{-6}\) by this value to find \(\tau\). The time constant \(\tau\) is approximately \(3.563 \times 10^{-6} \text{ seconds}\).
06

Finding the Capacitance \(C\)

Using \(\tau = RC\) and the given resistance \(R = 15.0 \times 10^3 \Omega\), solve for \(C\): \(C = \frac{\tau}{R}\). Substitute the calculated \(\tau\) and \(R\) to find \(C\).
07

Calculate \(C\)

Plug \(\tau = 3.563 \times 10^{-6}\,\text{s}\) and \(R = 15.0 \times 10^3 \Omega\) into \(C = \frac{\tau}{R}\). Thus, \(C \approx 2.375 \times 10^{-10} \text{ F}\) or \(237.5 \text{ pF}\).

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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, represented by the symbol \( \tau \), is crucial for understanding how quickly a capacitor charges through a resistor. This constant is calculated using the formula \( \tau = RC \), where \( R \) is the resistance in the circuit, and \( C \) is the capacitance of the capacitor.

The time constant \( \tau \) is a measure of the time it takes for the voltage across the capacitor to reach approximately 63.2% of its maximum value during charging. It also represents the time taken for the capacitor to discharge to about 36.8% of its initial charge when discharging.

Understanding the time constant helps in analyzing how quickly circuits respond to changes. For example, in the given exercise, the time constant is crucial in predicting when the voltage across the capacitor reaches 5V from a 12V supply.
Capacitor Charging Formula
During charging, a capacitor reaches a voltage \( V(t) \) at time \( t \) described by the formula: \( V(t) = V_0(1 - e^{-t/\tau}) \). Here, \( V_0 \) is the initial potential difference, and \( \tau \) is the time constant.

This equation tells us the voltage depends on the time and helps identify how a capacitor charges over time. Initially, the voltage rises rapidly, and as it approaches its maximum value, the rate of increase diminishes.

The formula results from integrating the basic charging equation of capacitors, illustrating an exponential relationship. In the exercise, the capacitor charging formula is used to solve how the voltage grows from 0V to 5V in \( 1.30 \mu s \) of time.
Capacitance Calculation
Capacitance \( C \), represented in Farads (F), measures a capacitor's ability to store charge per unit voltage. Calculating capacitance in an RC circuit involves using the known time constant and resistance: \( C = \frac{\tau}{R} \).

By rearranging the formula for time constant \( \tau = RC \), we can solve for \( C \) if both \( \tau \) and \( R \) are known.

In the exercise, after determining the time constant \( \tau \approx 3.563 \times 10^{-6} \text{s} \) and using the given resistance \( R = 15.0 \times 10^3 \Omega \), substitution into the formula gives \( C \approx 2.375 \times 10^{-10} \text{ F} \) or \( 237.5 \text{ pF} \). This calculation shows the effectiveness of capacitance in the circuit's behavior.

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Most popular questions from this chapter

An initially uncharged capacitor \(C\) is fully charged by a device of constant emf \(\mathscr{E}\) connected in series with a resistor \(R\) (a) Show that the final energy stored in the capacitor is half the energy supplied by the emf device. (b) By direct integration of \(i^{2} R\) over the charging time, show that the thermal energy dissipated by the resistor is also half the energy supplied by the emf device.

A capacitor with initial charge \(g_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

A group of \(N\) identical batteries of \(\operatorname{emf} \mathscr{E}\) and internal resistance \(r\) may be connected all in series (Fig. \(27-80 a\) ) or all in parallel (Fig. \(27-80 b\) ) and then across a resistor \(R\). Show that both arrangements give the same current in \(R\) if \(R=r\).

A car battery with a \(12 \mathrm{~V}\) emf and an internal resistance of \(0.040\) \(\Omega\) is being charged with a current of 50 A. What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{r}\) of energy dissipation inside the battery, and (c) the rate \(P_{\text {emr }}\) of energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) \(V\) and (e) \(P_{r}\) ?

A simple ohmmeter is made by connecting a \(1.50 \mathrm{~V}\) flashlight battery in series with a resistance \(R\) and an ammeter that reads from 0 to \(1.00 \mathrm{~mA}\), as shown in Fig. \(27-59 .\) 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) \(10.0 \%,(\mathrm{~b})\) \(50.0 \%\), 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 \(R ?\)
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