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Chapter 26: Problem 11

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?

Short Answer

Expert verified
Answer: When a single capacitor in an RC circuit is replaced with two identical capacitors connected in series, the time constant for the circuit decreases by half.

Step by step solution

01

Understand the RC circuit and the time constant

An RC circuit is a simple electrical circuit that contains a resistor (R) and a capacitor (C) connected in series. The time constant (蟿) of an RC circuit represents the time required for the voltage across the capacitor to either charge or discharge to about 63.2% of its final value. It is given by the equation: 蟿 = RC.
02

Determine the original time constant

Let's consider the original capacitor has capacitance C1, and the resistance of the circuit is R. The time constant for the original circuit, denoted as 蟿1, can be calculated using the formula 蟿 = RC. Thus, 蟿1 = RC1.
03

Find the equivalent capacitance for two capacitors in series

According to the problem, we replace the original capacitor with two identical capacitors connected in series. Let's denote their capacitances as C2 and C3. Since these capacitors are identical, C2 = C3 = C1. The formula for calculating the equivalent capacitance (Ceq) for capacitors connected in series is given by: 1/Ceq = 1/C2 + 1/C3 Plugging in the values of C2 and C3, we get: 1/Ceq = 1/C1 + 1/C1 = 2/C1 Now, we can find the equivalent capacitance: Ceq = C1/2
04

Calculate the new time constant

Now that we have the equivalent capacitance for the two capacitors connected in series, we can calculate the new time constant (蟿2) using the same formula as before, 蟿 = RC. However, this time we will use the equivalent capacitance, Ceq: 蟿2 = R * (C1/2) = (1/2)RC1
05

Compare the new time constant with the original time constant

To find the effect of replacing the single capacitor with two identical capacitors in series on the time constant, we need to compare the new time constant (蟿2) with the original time constant (蟿1): 蟿1 = RC1 蟿2 = (1/2)RC1 From this comparison, we can conclude that the new time constant is half of the original time constant. So, when the capacitor in an RC circuit is replaced with two identical capacitors connected in series, the time constant for the circuit decreases by half.

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Most popular questions from this chapter

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.)

Two capacitors in series are charged through a resistor. Identical capacitors are instead connected in parallel and charged through the same resistor. How do the times required to fully charge the two sets of capacitors compare?

You want to make an ohmmeter to measure the resistance of unknown resistors. You have a battery with voltage \(\mathrm{V}_{\mathrm{emf}}=9.00 \mathrm{~V}\), a variable resistor, \(R,\) and an ammeter that measures current on a linear scale from 0 to \(10.0 \mathrm{~mA}\) a) What resistance should the variable resistor have so that the ammeter gives its full-scale (maximum) reading when the ohmmeter is shorted? b) Using the resistance from part (a), what is the unknown resistance if the ammeter reads \(\frac{1}{4}\) of its full scale?

A battery has \(V_{\text {emf }}=12.0 \mathrm{~V}\) and internal resistance \(r=1.00 \Omega\). What resistance, \(R,\) can be put across the battery to extract \(10.0 \mathrm{~W}\) of power from it?

Which of the following will reduce the time constant in an RC circuit? a) increasing the dielectric constant of the capacitor b) adding an additional \(20 \mathrm{~m}\) of wire between the capacitor and the resistor c) increasing the voltage of the battery d) adding an additional resistor in parallel with the first resistor e) none of the above
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