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According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capacitance of \(32.0 \mu \mathrm{F}\) between two points \(A\) and \(B .\) (a) When one circuit is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance \(34.8 \mu \mathrm{F}\). To meet the specification, one additional capacitor can be placed between the two points. Should it be in series or in parallel with the \(34.8-\mu \mathrm{F}\) capacitor? What should be its capacitance? (b) What If? The next circuit comes down the assembly line with capacitance \(29.8 \mu \mathrm{F}\) between \(A\) and \(B .\) What additional capacitor should be installed in series or in parallel in that circuit, to meet the specification?

Short Answer

Expert verified
To meet the specification, an additional capacitor of \(C_2 \approx 186 \mu F\) should be installed in series with the \(34.8 \mu F\) capacitor, and an additional capacitor of \(C_2 \approx 2.2 \mu F\) should be installed in parallel with the \(29.8 \mu F\) capacitor.

Step by step solution

01

Reducing Capacitance

Adding a capacitor in series always results in a lower equivalent capacitance. So, to meet the specification, an additional capacitor must be installed in series with the \(34.8 \mu F\) capacitor. The formula for adding capacitors in series is \(1/C_{eq} = 1/C_1 + 1/C_2\). In this scenario, we know \(C_{eq} = 32 \mu F\) and \(C_1 = 34.8 \mu F\), we need to find \(C_2\). Rearranging the formula, we get \(C_2 = 1/ ( 1/C_{eq} - 1/C_1 )\). Plug the numbers and we find \(C_2 \approx 186 \mu F\).
02

Increasing Capacitance

Adding a capacitor in parallel always results in a greater equivalent capacitance. So, to meet the specification, an additional capacitor must be installed in parallel with the \(29.8 \mu F\) capacitor. The formula for adding capacitors in parallel is \(C_{eq} = C_1 + C_2\). In this scenario, we know \(C_{eq} = 32 \mu F\) and \(C_1 = 29.8 \mu F\), we need to find \(C_2\). Rearranging the formula, we get \(C_2 = C_{eq} - C_1\). Plug the numbers and we find \(C_2 \approx 2.2 \mu F\).

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Key Concepts

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

Series Capacitors
When capacitors are connected in series, their overall capacitance decreases. This is because the charge must travel through each capacitor sequentially in the circuit, which effectively lengthens the path the charge takes. Hence, having capacitors in series reduces the ability to store charge, resulting in lesser total capacitance.

To calculate the total or equivalent capacitance of capacitors in series, use the formula:
  • \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \)
Where \(C_{eq}\) is the equivalent capacitance and \(C_1, C_2, \ldots, C_n\) are the individual capacitances of the capacitors.

In the given exercise, a capacitor of \(34.8 \,\mu F\) is already in place, and a new one must be found that when in series gives a total of \(32 \,\mu F\). Using the rearranged formula, this can be solved to find the necessary capacitance of the new capacitor.
Parallel Capacitors
Capacitors connected in parallel result in an increase in total capacitance. This happens because each capacitor provides a separate path for charge to flow, which effectively doubles the amount of charge the circuit can hold. Thus, the total capacitance can increase dramatically with each capacitor added in parallel.

The formula for finding the equivalent capacitance for capacitors in parallel is quite straightforward:
  • \( C_{eq} = C_1 + C_2 + \ldots + C_n \)
Where \(C_{eq}\) is the total capacitance and \(C_1, C_2, \ldots, C_n\) are the individual capacitances.

In the context of the exercise, the initial capacitance measured is \(29.8 \,\mu F\) while the goal is \(32 \,\mu F\). By adding a capacitor in parallel, the equivalent capacitance can be increased to meet the desired specification. Solving the formula for the additional capacitance needed helps determine the size of this new capacitor.
Capacitance Calculation
Understanding and calculating capacitance is essential when working with capacitors in any electrical circuit. Capacitance is a measure of a capacitor’s ability to store charge per unit voltage and is measured in farads (\(F\)).

When solving such problems involving series and parallel configurations, it's important to determine whether the configuration calls for increasing or decreasing the capacitance. In a series configuration, the goal often involves reducing the total capacitance. Therefore, calculations focus on how a new series capacitor contributes to lowering the total. Conversely, in a parallel setup, the aim is often to enhance capacitance. Here, one calculates the added capacitance required to boost the total capacitance to a specified value.

Practicing these capacitance calculations in different configurations enables better circuit design and ensures devices work according to their design specifications. Computation tools like the series and parallel formulas allow us to adjust and optimize capacitance levels to suit specific needs, such as in the elevator timer circuit.

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Most popular questions from this chapter

Two capacitors when connected in parallel give an equivalent capacitance of \(9.00 \mathrm{pF}\) and give an equivalent capacitance of \(2.00 \mathrm{pF}\) when connected in series. What is the capacitance of each capacitor?

Two conductors having net charges of \(+10.0 \mu \mathrm{C}\) and \(-10.0 \mu \mathrm{C}\) have a potential difference of \(10.0 \mathrm{V}\) between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to \(+100 \mu \mathrm{C}\) and \(-100 \mu \mathrm{C} ?\)

An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \mathrm{cm}^{2},\) separated by a distance of 1.80 mm. A 20.0-V potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?

Two capacitors, \(C_{1}=25.0 \mu \mathrm{F}\) and \(C_{2}=5.00 \mu \mathrm{F},\) are connected in parallel and charged with a \(100-\mathrm{V}\) power supply. (a) Draw a circuit diagram and calculate the total energy stored in the two capacitors. (b) What If? What potential difference would be required across the same two capacitors connected in series in order that the combination stores the same amount of energy as in (a)? Draw a circuit diagram of this circuit.

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