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Chapter 17: Problem 63

Seven capacitors each of capacitance \(2 \mu \mathrm{F}\) are to be connected to obtain a capacitance of \((10 / 11) \mu \mathrm{F}\). Which of the following combinations is possible? (a) 5 in parallel, 2 in series (b) 4 in parallel, 3 in series (c) 3 in parallel, 4 in series (d) 2 in parallel, 5 in series

Short Answer

Expert verified
Option (a) 5 in parallel, 2 in series is correct.

Step by step solution

01

Understanding Capacitors in Series and Parallel

In parallel, the total capacitance is the sum of individual capacitances: \(C_{parallel} = C_1 + C_2 + \ldots + C_n\). In series, the reciprocal of the total capacitance is the sum of the reciprocals of individual capacitances: \(\frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}\).
02

Evaluate Option (a)

For 5 in parallel, \(C_{parallel} = 5 \times 2\, \mu F = 10\, \mu F\). With 2 in series, \(\frac{1}{C_{series}} = \frac{1}{2} + \frac{1}{2} = 1\), so \(C_{series} = 1\, \mu F\). Total \(C = \frac{10 \times 1}{10 + 1} = \frac{10}{11} \mu F\), which matches the required capacitance.
03

Evaluate Option (b)

For 4 in parallel, \(C_{parallel} = 4 \times 2\, \mu F = 8\, \mu F\). With 3 in series, \(\frac{1}{C_{series}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}\), so \(C_{series} = \frac{2}{3}\, \mu F\). Total \(C = \frac{8 \times \frac{2}{3}}{8 + \frac{2}{3}}\) which does not equal \(\frac{10}{11} \mu F\).
04

Evaluate Option (c)

For 3 in parallel, \(C_{parallel} = 3 \times 2\, \mu F = 6\, \mu F\). With 4 in series, \(\frac{1}{C_{series}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2\), so \(C_{series} = \frac{1}{2}\, \mu F\). Total \(C = \frac{6 \times \frac{1}{2}}{6 + \frac{1}{2}}\) which does not equal \(\frac{10}{11} \mu F\).
05

Evaluate Option (d)

For 2 in parallel, \(C_{parallel} = 2 \times 2\, \mu F = 4\, \mu F\). With 5 in series, \(\frac{1}{C_{series}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{5}{2}\), so \(C_{series} = \frac{2}{5}\, \mu F\). Total \(C = \frac{4 \times \frac{2}{5}}{4 + \frac{2}{5}}\) which does not equal \(\frac{10}{11} \mu F\).

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Key Concepts

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

Capacitors in Series
When capacitors are connected in series, they form a single path for the electric charge to flow. This setup might feel similar to setting up a series of dominoes 鈥� you tip one over, and the rest follow. In a series, the total capacitance is less than any single capacitor in the chain. The formula for capacitors in series is given by: \[ \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \]This reciprocal relationship means the total capacitance decreases as the number of capacitors in series increases. It's useful to remember this concept, for example, when aiming to achieve a precise capacitance using large, available capacitors instead of small ones. To calculate the total capacitance in series, you need to find the reciprocal of the sum of the reciprocals of all individual capacitances.
Capacitors in Parallel
Imagine capacitors in parallel like parallel lanes on a highway 鈥� more lanes mean more space for charge to flow, increasing the total capacitance. When capacitors are connected in parallel, the total capacitance is a straightforward sum of all individual capacitances. The formula is:\[ C_{\text{parallel}} = C_1 + C_2 + \ldots + C_n \]This setup effectively adds up the surface area of each capacitor's plates, which allows the device to store more charge. Larger capacitance values help in stabilizing voltage fluctuations, acting almost like a buffer. This is why parallel connections are common in circuits that need a stable electrical environment. Always add up the capacitances directly to find the total capacitance when connected in parallel.
Total Capacitance Formula
Understanding how to calculate total capacitance, whether in series, parallel, or a mixed combination, is a crucial skill. The total capacitance depends on how the capacitors are arranged. For complex circuits featuring both series and parallel capacitors: - First, identify and solve the parallel sections using the sum formula. - Next, apply the series formula for the series parts. - If there's a mix of both, handle each type of connection based on their position in the circuit. For a combination as seen in the example exercise: 1. Calculate the effective capacitance for parallel connections using the parallel capacitor formula. 2. Use the series capacitor formula for series connections of any portions of the circuit. 3. Finally, combine these results strategically to determine the total capacitance. Remember, combining different models of capacitors (series and parallel) strategically can help achieve required electrical values for complex technical tasks. This flexibility is essential for optimizing circuit designs, ensuring efficiency, and achieving desired electrical outcomes.

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Most popular questions from this chapter

In order to increase the capacity of a parallel plate condenser one should introduce between the plates a sheet of (a) mica (b) tin (c) copper (d) stainless steel

Suppose \(n\) identical capacitors are joined in parallel and charged to potential \(V .\) Now, they are separated and joined in series. If the energy possessed by each capacitor is \(U\), then on joining them in series, the energy and potential difference for the combination are (a) \(n U, V\) (b) \(U, n V\) (c) \(n U, n V\) (d) less than \(n U, n V\)

A variable parallel plate capacitor and an electroscope are connected in parallel to a battery. The reading of the electroscope would be decreased by (i) increasing the area of overlap of the plates (ii) placing a block of paraffin wax between the plates (iii) decreasing the distance between the plates (iv) decreasing the battery potential Then state if (a) only (i), (ii) and (iii) are correct (b) only (i) and (iii) are correct (c) only (ii) and (iv) are correct (d) only (iv) is correct

Two parallel metal plates carry charges \(+Q\) and \(-Q\) respectively. A test charge \(q_{o}\) placed between them experiences a force \(F\). Now the separation between the plates is doubled, then the force on the test charge will be (a) \(\bar{F}\) (b) \(2 F\) (c) \(F / 2\) (d) \(F / 4\)

A dielectric is placed in between the two parallel plates of a capacitor as shown in figure, the dielectric constant of the dielectric being \(K\). If the initial capacity is \(C\), then the new capacity will be (a) \(K C\) (b) \((K+1) C\) (c) \(C(K+1) / 2\) (d) \((K-1) C\)
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