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There are 10 condensers each of capacity \(5 \mu \mathrm{F}\). The ratio between maximum and minimum capacities obtained from these condensers will be (A) \(40: 1\) (B) \(25: 5\) (C) \(60: 3\) (D) \(100: 1\)

Short Answer

Expert verified
The maximum capacity is achieved by connecting all the condensers in parallel, resulting in \(50 \mu F\). The minimum capacity is obtained by connecting all the condensers in series, resulting in \(\frac{1}{2} \mu F\). The ratio between these capacities is \(\frac{50 \mu F}{\frac{1}{2} \mu F} = 100:1\), making the answer option (D).

Step by step solution

01

Find the maximum capacity

To find the maximum capacity, we can connect all the condensers in parallel. In this case, the total capacity is the sum of the individual capacities. As each condenser has a capacity of \(5 \mu F\), the total capacity for all 10 condensers connected in parallel will be: \(C_{max} = 10 \times 5 \mu F = 50 \mu F\)
02

Find the minimum capacity

To find the minimum capacity, we can connect all the condensers in series. In this case, the reciprocal of the total capacity is equal to the sum of the reciprocals of the individual capacities. Thus, \(\frac{1}{C_{min}} = \sum\limits^{10}_{i=1} \frac{1}{C_i}\) As each condenser has a capacity of \(5 \mu F\), we can write: \(\frac{1}{C_{min}} = 10 \times \frac{1}{5 \mu F} = \frac{10}{5 \mu F}\) Now, we can find the minimum capacity by taking the reciprocal of the previous expression: \(C_{min} = \frac{1}{\frac{10}{5 \mu F}} = \frac{5 \mu F}{10} = \frac{1}{2} \mu F\)
03

Find the ratio between maximum and minimum capacities

Now that we have the maximum and minimum capacities, we can determine the ratio between them: Ratio = \(\frac{C_{max}}{C_{min}} = \frac{50 \mu F}{\frac{1}{2} \mu F} = 100 \) Thus, the ratio between maximum and minimum capacities obtained from these condensers is 100:1, which corresponds to option (D).

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Key Concepts

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

Parallel Connection
When capacitors are connected in a parallel configuration, their individual capacitances simply add up. This setup is useful because it allows for the overall capacitance of the circuit to be increased. Think of parallel connection like having multiple paths in a river; water (or electrical charge, in this case) flows through all paths simultaneously and accumulates at the end.
There are a few key points to remember:
  • The total capacitance ( C_{total} ) is equal to the sum of the individual capacitances ( C_1, C_2,..., C_n ), written as C_{total} = C_1 + C_2 + ... + C_n .
  • All capacitors share the same voltage across them, making parallel connections ideal for situations where consistent voltage is needed.

In the exercise, connecting the capacitors in parallel yields the maximum capacitance because they all add up to increase the total capacity of the system.
Series Connection
In a series connection, capacitors are linked end-to-end so the charge flows sequentially from one capacitor to the next. This contrasts with a parallel setup and results in the total capacitance being less than the smallest individual capacitance in the chain!
Here's what you need to keep in mind about series connections:
  • The reciprocal of the total capacitance (C_{total}) is the sum of the reciprocals of each individual capacitor's capacitance, denoted as \( \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n} \).
  • This setup reduces the overall capacitance but can increase the voltage each capacitor can handle.

By placing all capacitors in the original exercise in series, we obtained the minimum possible capacitance of the configuration.
Capacity Calculation
Calculating the total capacitance of capacitors both in series and parallel connections is essential for designing circuits that function correctly.
In a parallel connection, as discussed, the process is straightforward—just sum up the capacitances. However, in a series configuration, you need to sum the reciprocals of each capacitor's capacitance first and then take the reciprocal of that sum to get the total capacitance.
Key formulae include:
  • Parallel: C_{total} = C_1 + C_2 + ... + C_n
  • Series: \( \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n} \)

Remember to use these strategies to find the maximum and minimum capacities, which were calculated in steps 1 and 2 of the original solution.
Electrical Circuits
Electrical circuits are the systems through which electricity flows. They consist of various components such as resistors, capacitors, and wires through which current moves.
Circuits can vary from simple, with just one or two components, to complex arrays that perform all manner of tasks. Understanding how each element contributes to the overall function of the circuit is crucial in circuit design and analysis.
Capacitors in circuits serve functions like stabilizing voltage and filtering signals, essential in applications ranging from basic electronic devices to advanced technological systems.
In the exercise, understanding the principles of capacitors in both series and parallel connections provided insights into how these configurations affect total capacitance and voltage handling capabilities. The principles of electrical circuitry help in adeptly manipulating such configurations to achieve desired performance outcomes.

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Most popular questions from this chapter

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(\ell\). The maximum possible flux of the electric field through the cube will be ....... (A) \(\sqrt{3}\left(\lambda \ell / \in_{0}\right)\) (B) \(\left(\lambda \ell / \in_{0}\right)\) (C) \(\sqrt{2}\left(\lambda \ell / \in_{0}\right)\) (D) \(\left[\left(6 \lambda \ell^{2}\right) / \epsilon_{0}\right]\)

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) \(\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]\) (B) \(\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]\) (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)

In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d. \(2400 \mathrm{v}\) between the plates. To keep a drop of half the radius stationary the potential difference had to be made \(600 \mathrm{v}\). What is the charge on the second drop? (A) \([(3 Q) / 2]\) (B) \((\mathrm{Q} / 4)\) (C) \(Q\) (D) \((\mathrm{Q} / 2)\)

Electric potential at any point is \(\mathrm{V}=-5 \mathrm{x}+3 \mathrm{y}+\sqrt{(15 \mathrm{z})}\), then the magnitude of the electric field is \(\ldots \ldots \ldots \mathrm{N} / \mathrm{C}\). (A) \(3 \sqrt{2}\) (B) \(4 \sqrt{2}\) (C) 7 (D) \(5 \sqrt{2}\)

The plates of a parallel capacitor are charged up to \(100 \mathrm{~V}\). If \(2 \mathrm{~mm}\) thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by \(1.6 \mathrm{~mm}\) the dielectric constant of the plate is (A) 5 (B) 4 (C) \(1.25\) (D) \(2.5\)

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