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Chapter 26: Problem 13

Explain why the time constant for an \(\mathrm{RC}\) circuit increases with \(R\) and with \(C\). (The answer "That's what the formula says" is not sufficient.)

Short Answer

Expert verified
Answer: The time constant for an RC circuit increases with resistance and capacitance because a larger resistance slows down the flow of charge in the circuit, while a larger capacitance requires more time to charge or discharge due to its increased charge storage capacity. This is reflected in the formula for the time constant, which is the product of resistance and capacitance: 蟿 = R 脳 C.

Step by step solution

01

Defining the time constant of an RC circuit

The time constant, denoted as 蟿 (tau), of an RC circuit is a measure of the time it takes for the voltage across the capacitor to charge (or discharge) to approximately 63.2% of its final value when a step voltage is applied across the circuit. The time constant is crucial in determining the speed at which an RC circuit responds to changes in voltage.
02

Formula for time constant

The time constant for an RC circuit is given by the product of the resistance (R) and the capacitance (C) of the circuit: \(\tau = R \times C\). This formula shows that the time constant is directly proportional to both the resistance and the capacitance.
03

Understanding the role of resistance in an RC circuit

The resistance in an RC circuit determines how easily charge flows through the circuit. A larger resistance will impede the flow of charge more than a smaller resistance. This means that, for a given capacitance, it takes more time for the capacitor to charge (or discharge) when the resistance is high.
04

Understanding the role of capacitance in an RC circuit

The capacitance of an RC circuit determines how much charge can be stored on the capacitor for a given voltage. A larger capacitance means that more charge can be stored, and vice versa. Hence, for a given resistance in the circuit, it will take more time for a capacitor with a larger capacitance to charge (or discharge) because it has to store more charge.
05

Combining the effects of resistance and capacitance on the time constant

When resistance and capacitance both increase, their combined effect on the time constant is also increased. Since the time constant is the product of resistance and capacitance, an increase in either of these values will result in an increase in the time constant. This is consistent with our understanding that a larger resistance slows down the flow of charge in the circuit, while a larger capacitance requires more time to charge or discharge due to its increased charge storage capacity.

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Most popular questions from this chapter

A capacitor \(C\) is initially uncharged. At time \(t=0,\) the capacitor is attached through a resistor \(R\) to a battery. The energy stored in the capacitor increases, eventually reaching a value \(U\) as \(t \rightarrow \infty\), After a time equal to the time constant \(\tau=R C\), the energy stored in the capacitor is given by a) \(U / e\). c) \(U(1-1 / e)^{2}\) b) \(U / e^{2}\) d) \(U(1-1 / e)\).

Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=\) 0 , the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega\), that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q\) ?

A parallel plate capacitor with \(C=0.050 \mu \mathrm{F}\) has a separation between its plates of \(d=50.0 \mu \mathrm{m} .\) The dielectric that fills the space between the plates has dielectric constant \(\kappa=2.5\) and resistivity \(\rho=4.0 \cdot 10^{12} \Omega \mathrm{m} .\) What is the time constant for this capacitor? (Hint: First calculate the area of the plates for the given \(C\) and \(\kappa\), and then determine the resistance of the dielectric between the plates.)

A resistor and a capacitor are connected in series. If a second identical capacitor is connected in series in the same circuit, the time constant for the circuit will a) decrease. b) increase. c) stay the same.

A \(12.0-V\) battery is attached to a \(2.00-\mathrm{mF}\) capacitor and a \(100 .-\Omega\) resistor. Once the capacitor is fully charged, what is the energy stored in it? What is the energy dissipated as heat by the resistor as the capacitor is charging?
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