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Chapter 26: Problem 24

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鈥檚 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

A parallel-plate capacitor has a charge \(Q\) and plates of area \(A\). What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is \(E=Q / A \epsilon_{0},\) you might think that the force is \(F=Q E=Q^{2} / A \epsilon_{0} .\) This is wrong, because the field \(E\) includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually \(F=Q^{2} / 2 \epsilon_{0} A .\) (Suggestion: Let \(C=\epsilon_{0} A / x\) for an arbitrary plate separation \(x ;\) then require that the work done in separating the two charged plates be \(W=\int F \, d x\).) The force exerted by one charged plate on another is sometimes used in a machine shop to hold a workpiece stationary.

A detector of radiation called a Geiger tube consists of a closed, hollow, conducting cylinder with a fine wire along its axis. Suppose that the internal diameter of the cylinder is \(2.50 \mathrm{cm}\) and that the wire along the axis has a diameter of \(0.200 \mathrm{mm} .\) The dielectric strength of the gas between the central wire and the cylinder is \(1.20 \times 10^{6} \mathrm{V} / \mathrm{m} .\) Calculate the maximum potential difference that can be applied between the wire and the cylinder before breakdown occurs in the gas.

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?

A capacitor is constructed from two square plates of sides \(\ell\) and separation \(d\), as suggested in Figure \(\mathrm{P} 26.64 .\) You may assume that \(d\) is much less than \(\ell\). The plates carry charges \(+Q_{0}\) and \(-Q_{0} .\) A block of metal has a width \(\ell\), a length \(\ell\) and a thickness slightly less than \(d\). It is inserted a distance \(x\) into the capacitor. The charges on the plates are not disturbed as the block slides in. In a static situation, a metal prevents an electric field from penetrating inside it. The metal can be thought of as a perfect dielectric, with \(\kappa \rightarrow \infty,\) (a) Calculate the stored energy as a function of \(x\) (b) Find the direction and magnitude of the force that acts on the metallic block. (c) The area of the advancing front face of the block is essentially equal to \(\ell d\). Considering the force on the block as acting on this face, find the stress (force per area) on it. (d) For comparison, express the energy density in the electric field between the capacitor plates in terms of \(Q_{0}, \ell, d,\) and \(\epsilon_{0}\).

The immediate cause of many deaths is ventricular fibrillation, uncoordinated quivering of the heart as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A defibrillator (Fig. 26.14 ) is a device that applies a strong electric shock to the chest over a time interval of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about \(8 \mathrm{cm}\) across and coated with conducting paste, are held against the chest on both sides of the heart. Their handles are insulated to prevent injury to the operator, who calls, "Clear!" and pushes a button on one paddle to discharge the capacitor through the patient's chest. Assume that an energy of \(300 \mathrm{J}\) is to be delivered from a \(30.0-\mu \mathrm{F}\) capacitor. To what potential difference must it be charged?
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