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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 \(n U, n V\)

Short Answer

Expert verified
(c) nU, nV

Step by step solution

01

Identify Initial Capacitor Configuration

Initially, we have \(n\) identical capacitors each of capacitance \(C\) connected in parallel and charged to a potential \(V\). In this configuration, the total capacitance of the parallel combination is the sum of all individual capacitance, which is \(nC\).
02

Calculate Energy in Parallel Configuration

The energy stored in a capacitor is given by \(U = \frac{1}{2}CV^2\). For a single capacitor, the energy is \(U = \frac{1}{2}CV^2\). Since all capacitors are identical, in the parallel configuration, each capacitor has energy \(U\).
03

Reconfigure Capacitors to Series

After separating and reconfiguring the capacitors in series, the equivalent capacitance \(C_{eq}\) of the series combination becomes \(\frac{C}{n}\).
04

Calculate Voltage for Series Configuration

The potential difference across the series combination of \(n\) capacitors is the sum of individual voltages across each capacitor. In a series setup, the total potential is \(nV\).
05

Calculate Energy in Series Configuration

The energy for the series configuration is given by \(U' = \frac{1}{2}C_{eq}(nV)^2 = \frac{1}{2}\left(\frac{C}{n}\right)(nV)^2 = \frac{1}{2}CnV^2 = nU\).
06

Compare Options with Final Calculations

The calculated energy for series combination is \(nU\), and the potential difference is \(nV\). Hence, the correct answer is \(c)\, nU, nV\).

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Key Concepts

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

Energy Stored in Capacitors
When we talk about capacitors, one of the key features is their ability to store energy. This stored energy is essential because it allows capacitors to release energy quickly, which is useful in many electronic circuits. The energy stored in a capacitor can be calculated using the formula:
\[U = \frac{1}{2} C V^2\]
where:
  • \(U\) is the energy stored in the capacitor.
  • \(C\) is the capacitance of the capacitor.
  • \(V\) is the potential difference (voltage) across the capacitor.
The formula shows that the energy is proportional to the capacitance and the square of the voltage. This means that even a small increase in voltage can lead to a significant increase in stored energy. Understanding this relationship is essential when designing circuits that require precise energy management.
Capacitors in Series and Parallel
Capacitors can be connected in different configurations, mainly in series or parallel, each affecting the overall circuit differently.

  • **Parallel Configuration:** In a parallel arrangement, each capacitor is connected directly across the same two points, providing the same voltage to each. The total capacitance is simply the sum of all individual capacitances:
    \[C_{\text{total}} = C_1 + C_2 + \ldots + C_n\]This configuration means more stored charge because the equivalent capacitance is higher.
  • **Series Configuration:** When capacitors are in series, they connect end to end, creating a single path for charge flow. The total capacitance is less than any individual capacitor's capacitance and is given by:
    \[\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}\]This reduces the equivalent capacitance, increasing the total voltage each stores but decreasing the overall capacitance.
Being confident about these differences ensures you select the correct configuration for your desired circuit performance.
Equivalent Capacitance
Understanding equivalent capacitance is crucial in simplifying complex circuits into a single capacitor equivalent to the actual configuration of multiple capacitors. This process makes analyzing and understanding complex circuits more manageable.

In a **parallel configuration**, the equivalent capacitance increases as more capacitors are added. This is because all capacitors contribute directly to the total capacitance. Imagine stacking containers of energy, where adding more increases the overall capacity to store energy.

Conversely, in a **series configuration**, the equivalent capacitance decreases as more capacitors are added. Think of a single barrier across multiple blocks, where the overall ability to store energy decreases because each capacitor adds a slight resistance to the flow of energy.

Whether you need to increase total energy storage or adjust how energy is managed across a potential difference, mastering these concepts helps in designing circuits optimized for specific tasks, ensuring efficient functioning and reliability.

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Most popular questions from this chapter

Force acting upon a charged particle kept between the plates of a charged capacitor is \(F\). If one of the plates of the capacitor is removed, force acting on the same particle will becomes (a) 0 (b) \(F / 2\) (c) \(F\) (d) \(2 F\)

An uncharged parallel plate capacitor having a dielectric of constant \(K\) is connected to a similar air-cored parallel capacitor charged to a potential \(V .\) The two share the charge and the common potential is \(V^{\prime} .\) the dielectric constant \(\mathrm{K}\) is (a) \(\frac{V^{\prime}-V}{V^{\prime}+V}\) (b) \(\frac{V^{\prime}-V}{V^{\prime}}\) (c) \(\frac{V^{\prime}-V}{V}\) (d) \(\frac{V-V^{\prime}}{V}\)

Two spherical conductors \(A\) and \(B\) of radii \(a\) and \(b\) \((b>a)\) are placed concentrically in air. \(B\) is given a charge \(+Q\) and \(A\) is earthed. The equivalent capacitance of the system is (a) \(4 \pi \varepsilon_{o}\left(\frac{a b}{b-a}\right)\) (b) \(4 \pi \varepsilon_{e}(a+b)\) (c) \(4 \pi \varepsilon_{o} b\) (d) \(4 \pi \varepsilon_{o}\left(\frac{b^{2}}{b-a}\right)\)

The plates of a capacitor are charged to a potential difference of 320 volt and are then connected to a resistor. The potential difference across the capacitor decays exponentially with time. After 1 sec, the potential difference between the plates of the capacitor is \(240 \mathrm{~V}\), then after 2 and 3 seconds the potential difference between the plates will be, respectively (a) \(200 \mathrm{~V}\) and \(180 \mathrm{~V}\) (b) \(180 \mathrm{~V}\) and \(135 \mathrm{~V}\) (c) \(160 \mathrm{~V}\) and \(80 \mathrm{~V}\) (d) \(140 \mathrm{~V}\) and \(20 \mathrm{~V}\)

Two insulated charged spheres of radii \(R_{1}\) and \(R_{2}\) having charges \(Q_{1}\) and \(Q_{2}\) respectively are connected to each other, then there is (a) no change in the energy of the system (b) an increase in the energy of the system (c) always a decrease in the energy of the system (d) a decrease in energy of the system unless \(q_{1} R_{2}=q_{2} R_{1}\)

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