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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 (τ) 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: τ = 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 τ1, can be calculated using the formula τ = RC. Thus, τ1 = 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 (τ2) using the same formula as before, τ = RC. However, this time we will use the equivalent capacitance, Ceq: τ2 = 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 (τ2) with the original time constant (τ1): τ1 = RC1 τ2 = (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

You wish to measure both the current through and the potential difference across some component of a circuit. It is not possible to do this simultaneously and accurately with ordinary voltmeters and ammeters. Explain why not.

Which of the following has the same unit as the electromotive force (emf)? a) current b) electric potential c) electric field d) electric power e) none of the above

Kirchhoff's Junction Rule states that a) the algebraic sum of the currents at any junction in a circuit must be zero. b) the algebraic sum of the potential changes around any closed loop in a circuit must be zero. c) the current in a circuit with a resistor and a capacitor varies exponentially with time. d) the current at a junction is given by the product of the resistance and the capacitance. e) the time for the current development at a junction is given by the product of the resistance and the capacitance.

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.)

How can you light a \(1.0-\mathrm{W}, 1.5-\mathrm{V}\) bulb with your \(12.0-V\) car battery?
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