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You have \(N\) identical capacitors, each with capacitance \(C\), connected in parallel. The equivalent capacitance of this system of capacitors is a) \(N C\). b) \(C / N\). c) \(N^{2} C\). d) \(C / N^{2}\) e) \(C\).

Short Answer

Expert verified
Answer: a) \(NC\)

Step by step solution

01

Analyzing the given answer choices

We have the following answer options: a) \(NC\), b) \(C/N\), c) \(N^2C\), d) \(C/N^2\), e) \(C\). Our aim is to find out which one of them represents the equivalent capacitance of N identical capacitors connected in parallel.
02

Equivalent capacitance of capacitors in parallel

In a parallel connection of capacitors, the equivalent capacitance is calculated as the summation of the individual capacitances. So, for N capacitors in parallel we have: \(C_{eq} = C_1 + C_2 + ... + C_N\) Since all capacitors are identical with capacitance C, we can write: \(C_{eq} = C + C + ... + C\) (N times)
03

Simplifying the equation

As we have N identical capacitors, their capacitance is same, summing them up gives: \(C_{eq} = NC\) Now let's compare our result with the answer choices given: Our result is \(C_{eq} = NC\), which matches with option (a). Therefore, the equivalent capacitance of N identical capacitors connected in parallel is \(NC\).

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Key Concepts

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

Capacitors in Parallel
When capacitors are connected in parallel, each one has the same voltage across it as the other capacitors in the parallel network. The key to understanding parallel circuits is recognizing that, unlike resistors in parallel, the total (or equivalent) capacitance increases.

Imagine having multiple storage tanks (capacitors) with the same size of opening (voltage) but different capacities (capacitance); hooking them up in parallel creates one big tank with the combined volume of all the smaller tanks. Applied to capacitors, this analogy illustrates that the charge stored in the combined system is the sum of charges that each capacitor would store individually if the same voltage was applied to them.

As we delve into the problem exercise, we consider that the equivalent capacitance of identical capacitors connected in parallel is the summation of their individual capacitances. This property stems from their ability to store electrical energy, and when capacitors are connected in parallel, they effectively increase the plate area available for storing charge, leading to a higher combined capacitance.
Capacitance Calculation
The formula for the equivalent capacitance, denoted as \(C_{eq}\), of capacitors in parallel takes on a very straightforward form due to the additive nature of capacitance in such an arrangement. You essentially add up the capacitance values directly without needing to take reciprocals or perform more complex operations as with resistors or capacitors in series.

For a set of \(N\) identical capacitors each with capacitance \(C\) in parallel, the equation simplifies elegantly into \(C_{eq} = NC\), representing a direct multiplication of the number of capacitors by the individual capacitance value. This simplicity can be a pitfall for students who might overthink the problem, expecting more complicated calculations like in series circuits. Remembering that parallel capacitor connections increase the system's ability to store charge can help prevent such confusion.
Series and Parallel Circuits
Series and parallel circuits differ mainly in how components are connected. In series circuits, components are connected end-to-end, which means the same current flows through each component, but the total voltage is divided among them. In parallel circuits, on the other hand, components are connected across the same two points, sharing the same voltage but allowing current to divide among them.

The major consequence of a parallel circuit in terms of capacitance is it being additive, while in series circuits, the reciprocal values are additive. This contrast between the behavior of series and parallel circuits in the electrical domain should be a fundamental tenet in the understanding of any student delving into electronics and electrical engineering.

Through our exercise example, it becomes evident that while the formula in series for calculating equivalent capacitance may involve division or taking reciprocals, in a parallel setup, it's the straight summation of capacitors' values that gives us the total capacitance, underscoring the importance of recognizing the nature of the circuit before starting the calculation.

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Most popular questions from this chapter

The battery of an electric car stores 67.39 MJ of energy. If 6845 supercapacitors, each with capacitance \(C\) and charged to a potential difference of \(2.377 \mathrm{~V}\), can supply this amount of energy, what is the value of \(C\) for each supercapacitor?

A proton traveling along the \(x\) -axis at a speed of \(1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters the gap between the plates of a \(2.00-\mathrm{cm}\) -wide parallel plate capacitor. The surface charge distributions on the plates are given by \(\sigma=\pm 1.00 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) How far has the proton been deflected sideways \((\Delta y)\) when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

Calculate the maximum surface charge distribution that can be maintained on any surface surrounded by dry air.

A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between them?

The capacitance of a spherical capacitor consisting of two concentric conducting spheres with radii \(r_{1}\) and \(r_{2}\left(r_{2}>r_{1}\right)\) is given by \(C=4 \pi \epsilon_{0} r_{1} r_{2} /\left(r_{2}-r_{1}\right) .\) Suppose that the space between the spheres, from \(r_{1}\) up to a radius \(R\left(r_{1}

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