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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: τ = R × C.

Step by step solution

01

Defining the time constant of an RC circuit

The time constant, denoted as τ (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

The ammeter your physics instructor uses for in-class demonstrations has internal resistance \(R_{\mathrm{i}}=75 \Omega\) and measures a maximum current of \(1.5 \mathrm{~mA}\). The same ammeter can be used to measure currents of much greater magnitudes by wiring a shunt resistor of relatively small resistance, \(R_{\text {shunt }}\), in parallel with the ammeter. (a) Sketch the circuit diagram, and explain why the shunt resistor connected in parallel with the ammeter allows it to measure larger currents. (b) Calculate the resistance the shunt resistor has to have to allow the ammeter to measure a maximum current of 15 A.

If the capacitor in an \(\mathrm{RC}\) circuit is replaced with two identical capacitors connected in series, what happens to the time constant for the circuit?

A circuit consists of two \(100 .-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the potential drop across one of the resistors. b) A voltmeter with internal resistance \(10.0 \mathrm{M} \Omega\) is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

How long will it take for the current in a circuit to drop from its initial value to \(1.50 \mathrm{~mA}\) if the circuit contains two \(3.8-\mu \mathrm{F}\) capacitors that are initially uncharged, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) battery all connected in series?

A circuit consists of a source of emf, a resistor, and a capacitor, all connected in series. The capacitor is fully charged. How much current is flowing through it? a) \(i=V / R\) b) zero c) neither (a) nor (b)
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