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You have three capacitors: \(C_{1}=67 \mu \mathrm{F}, C_{2}=45 \mu \mathrm{F},\) and \(C_{3}=33 \mu \mathrm{F}\) . Determine the maximum equivalent capacitance you can obtain by connecting two of the capacitors in parallel and then connecting the parallel combination in series with the remaining capacitor.

Short Answer

Expert verified
The maximum equivalent capacitance is 35.57 µF.

Step by step solution

01

Understand the Problem

We need to determine the maximum equivalent capacitance by making a combination of three capacitors where two are in parallel and one is in series with this pair. The equivalent capacitance will be calculated based on these connections.
02

Parallel Combination Formula

For capacitors in parallel, the equivalent capacitance is the sum of the individual capacitances. Therefore, for two capacitors in parallel, labeled as \(C_p\), it is given by:\[C_p = C_i + C_j\]where \(i\) and \(j\) are the indices of the capacitors chosen to be in parallel.
03

Calculate Each Parallel Combination

Calculate the equivalent capacitance for each pair:- For \(C_1\) and \(C_2\): \(C_p = 67 + 45 = 112 \mu F\).- For \(C_1\) and \(C_3\): \(C_p = 67 + 33 = 100 \mu F\).- For \(C_2\) and \(C_3\): \(C_p = 45 + 33 = 78 \mu F\).
04

Series Combination Formula

For capacitors in series, the equivalent capacitance \(C_s\) is given by the formula:\[\frac{1}{C_s} = \frac{1}{C_p} + \frac{1}{C_k}\]where \(C_k\) is the third capacitor not involved in the parallel combination.
05

Calculate Series Combinations

Using the parallel combinations from Step 3, compute the series equivalent capacitance:- For \(C_p = 112 \mu F\) (parallel of \(C_1\) and \(C_2\)) and \(C_3\), \(C_s\) is: \[ \frac{1}{C_s} = \frac{1}{112} + \frac{1}{33} \] \(C_s\) will approximately be \(25.54 \mu F\).- For \(C_p = 100 \mu F\) (parallel of \(C_1\) and \(C_3\)) and \(C_2\), \(C_s\) is: \[ \frac{1}{C_s} = \frac{1}{100} + \frac{1}{45} \] \(C_s\) will approximately be \(31.03 \mu F\).- For \(C_p = 78 \mu F\) (parallel of \(C_2\) and \(C_3\)) and \(C_1\), \(C_s\) is: \[ \frac{1}{C_s} = \frac{1}{78} + \frac{1}{67} \] \(C_s\) will approximately be \(35.57 \mu F\).
06

Choose Maximum Equivalent Capacitance

From our calculations:- \(C_s\) for (\(C_1 \) and \(C_2\) parallel) and \(C_3\): \(25.54 \mu F\)- \(C_s\) for (\(C_1 \) and \(C_3\) parallel) and \(C_2\): \(31.03 \mu F\)- \(C_s\) for (\(C_2 \) and \(C_3\) parallel) and \(C_1\): \(35.57 \mu F\)The maximum equivalent capacitance is \(35.57 \mu F\).

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Key Concepts

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

Parallel capacitors
When capacitors are connected in parallel, the total or equivalent capacitance increases. This is because, in a parallel configuration, the plates of the capacitors are connected side-by-side. Essentially, it's like increasing the surface area of one big capacitor. The result is a higher ability to store charge.
To calculate the total capacitance of capacitors in parallel, simply add up the capacitance values of all capacitors involved. The formula for the equivalent capacitance, denoted as \( C_p \), is straightforward:
  • \( C_p = C_1 + C_2 + C_3 + \ldots\)
In the context of the original exercise, various pairs of capacitors were connected in parallel to find which combination would yield the maximum equivalent capacitance when further paired with a series capacitor. The greater the equivalent capacitance, the better the ability to store more electric charge in the system.
Series capacitors
Capacitors in series work differently compared to parallel connections. When connected in series, the total capacitance of the system decreases. In this configuration, each capacitor shares the same charge, but the voltage is divided among them. It's akin to stretching the separation between plates, effectively reducing the system's capacity for storing charge.
The formula for equivalent capacitance \( C_s \) when capacitors are in series is based on the reciprocals of their individual capacitances:
  • \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots\)
The original exercise used this principle to find the equivalent capacitance when a parallel combination of capacitors was put in series with a third capacitor. Each configuration resulted in a different capacitance, and the challenge was to find the combination that gave the maximum series capacitance.
Equivalent capacitance
The concept of equivalent capacitance helps simplify complex electrical systems into simpler components. By determining the equivalent capacitance of a combination of capacitors, one can treat the whole set as if it were a single capacitor with that capacitance.
In circuits, the strategic arrangement of capacitors (in series or parallel) changes the total storage capacity of the system. The idea is to either increase the capacitance (using parallel configurations) or adjust it for specific circuit requirements (such as tuning or energy control, often modified using series).
For the given exercise, finding the maximum equivalent capacitance required calculating how different combinations of series and parallel connections affected the total capacitance. By doing this, electrical engineers ensure circuits perform optimally, whether storing energy efficiently or distributing it effectively.

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Most popular questions from this chapter

A defibrillator is used during a heart attack to restore the heart to its normal beating pattern (see Section 19.5). A defibrillator passes 18 A of current through the torso of a person in 2.0 ms. (a) How much charge moves during this time? (b) How many electrons pass through the wires connected to the patient?

According to Equation 20.7, an ac voltage \(V\) is given as a function of time \(t\) by \(V=V_{0} \sin 2 \pi f t,\) where \(V_{0}\) is the peak voltage and \(f\) is the frequency (in hertz). For a frequency of 60.0 \(\mathrm{Hz}\) , what is the smallest value of the time at which the voltage equals one-half of the peak value?

A coffee cup heater and a lamp are connected in parallel to the same \(120-\mathrm{V}\) outlet. Together, they use a total of 111 \(\mathrm{W}\) of power. The resistance of the heater is \(4.0 \times 10^{2} \Omega .\) Find the resistance of the lamp.

The coil of wire in a galvanometer has a resistance of \(R_{C}=60.0 \Omega\) . The galvanometer exhibits a full-scale deflection when the current through it is 0.400 \(\mathrm{mA}\) . A resistor is connected in series with this combination so as to produce a nondigital voltmeter. The voltmeter is to have a full-scale deflection when it measures a potential difference of 10.0 \(\mathrm{V}\) . What is the resistance of this resistor?

The current in a \(47-\Omega\) resistor is 0.12 A. This resistor is in series with a \(28-\Omega\) resistor, and the series combination is connected across a battery. What is the battery voltage?

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