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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?

Short Answer

Expert verified
Answer: The capacitors connected in parallel will take four times longer to fully charge compared to the capacitors connected in series.

Step by step solution

01

Understand Capacitor Charging through a Resistor

When a capacitor is connected to a voltage source through a resistor, the charging process can be described by the time constant τ, which is defined as the product of the resistance R and the capacitance C (τ=RC). The time constant determines how quickly the capacitor charges - a larger time constant means that the capacitor charges slower, while a smaller one means it charges faster.
02

Calculate Equivalent Capacitances for Series and Parallel Configurations

When capacitors are connected in series or parallel, their overall capacitance can be determined using the following formulas: For capacitors in series: 1/C_eq = 1/C_1 + 1/C_2 For capacitors in parallel: C_eq = C_1 + C_2 Since the capacitors are identical, we can simplify these formulas: For capacitors in series: 1/C_eq = 2/C C_eq = C/2 For capacitors in parallel: C_eq = 2*C
03

Calculate Time Constants for Capacitors in Series and Parallel

Next, we'll calculate the time constant for both series and parallel configurations using their equivalent capacitances and the resistor's resistance: For capacitors in series: τ_series = R * C_eq_series τ_series = R * (C/2) For capacitors in parallel: τ_parallel = R * C_eq_parallel τ_parallel = R * (2*C)
04

Compare the Time Constants

Now that we have the time constants for both configurations, we can compare the times required to fully charge the capacitors by analyzing their time constants: τ_series = R * (C/2) τ_parallel = R * (2*C) As τ_parallel is four times larger than τ_series (τ_parallel = 4 * τ_series), the capacitors connected in parallel will take four times longer to fully charge compared to the capacitors connected in series.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series and Parallel Capacitors
Understanding how capacitors are connected is crucial to predict how they will behave in a circuit. For capacitors in series, the charge on each capacitor is the same, and the total voltage is the sum of the voltages across each capacitor. The formula to calculate the equivalent capacitance (\( C_{eq} \) for capacitors in series is:

\[\[\begin{align*}\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdotsd{align}
Which for identical capacitors simplifies to:\[C_{eq} = \frac{C}{n}\].
In contrast, for parallel connections, each capacitor experiences the same voltage, but the charges can differ. Their total capacitance is the sum of their individual capacitances, making the formula for equivalent capacitance in parallel configuration simply:\[C_{eq} = C_1 + C_2 + C_3 + \cdots\].
For identical capacitors in parallel, this sums up to:\[C_{eq} = nC\], where 'n' is the number of capacitors. Thus, the behavior of capacitors in a circuit significantly differs depending on whether they are wired in series or parallel.
Time Constant (τ)
The time constant, represented by the Greek letter tau (τ), is a critical concept when discussing capacitor charging. It defines the speed at which a capacitor charges through a resistor. Mathematically, the time constant can be expressed as:\[\tau = R \times C\].
Here, 'R' represents the resistance, and 'C' is the capacitance in the circuit. The time constant is important because it tells us that after a period equal to one time constant, the capacitor will have charged to about 63.2% of its maximum voltage level. After five time constants, the capacitor is considered to be fully charged (over 99% of the full charge). Knowing the time constant gives us an idea of the charging speed - a smaller τ indicates faster charging, and a larger τ means a slower charging process.
Equivalent Capacitance
Equivalent capacitance (\( C_{eq} \)) comes into play when we have several capacitors combined in series or parallel in a circuit. It's a way of simplifying the circuit to a single capacitor that would have the same effect as the original combination. For instance, if two identical capacitors are connected in series, the equivalent capacitance is half of one of the capacitors:\[C_{eq}\_series = \frac{C}{2}\].
Alternatively, if they are connected in parallel, the equivalent capacitance is double:\[C_{eq}\_parallel = 2C\].
Understanding equivalent capacitance is crucial when determining the time required to fully charge a set of capacitors, as it directly influences the time constant τ of the circuit. In turn, this affects charging time - a circuit's response to a change in voltage essentially depends on its equivalent capacitance and total resistance.

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Most popular questions from this chapter

How long will it take, as a multiple of the time constant, \(\tau,\) for the capacitor in an RC circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)

A Wheatstone bridge is constructed using a \(1.00-\mathrm{m}\) -long Nichrome wire (the purple line in the figure) with a conducting contact that can slide along the wire. A resistor, \(R_{1}=100 . \Omega\), is placed on one side of the bridge, and another resistor, \(R\), of unknown resistance, is placed on the other side. The contact is moved along the Nichrome wire, and it is found that the ammeter reading is zero for \(L=25.0 \mathrm{~cm} .\) Knowing that the wire has a uniform cross section throughout its length, determine the unknown resistance.

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.)

An \(\mathrm{RC}\) circuit has a time constant of \(3.10 \mathrm{~s}\). At \(t=0,\) the process of charging the capacitor begins. At what time will the energy stored in the capacitor reach half of its maximum value?

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

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