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Three capacitors having capacitances 8.4, 8.4, and \(4.2 \mu \mathrm{F}\) are connected in series across a \(36.0-\mathrm{V}\) potential difference. (a) What is the total energy stored in all three capacitors? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged plates connected together. What is the total energy now stored in the capacitors?

Short Answer

Expert verified
The total energy stored in all three capacitors when they are connected in series is \(20.5\,mJ\). When the capacitors are reconnected in parallel, the total energy stored in the capacitors is \(486\,mJ\).

Step by step solution

01

Part A - Calculate the equivalent capacitance in series configuration

To find the equivalent capacitance when capacitors are connected in series, we can use the following formula: \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \) Where \(C_1 = 8.4\mu F\), \(C_2 = 8.4\mu F\), and \(C_3 = 4.2\mu F\).
02

Part A - Find the voltage across each capacitor in the series configuration

To find the voltage across each capacitor in the series configuration, we can use the formula: \(V_i = Q_i/C_i\) Where \(V_i\) is the voltage across the i-th capacitor, \(Q_i\) is the charge on the i-th capacitor, and \(C_i\) is the capacitance of the i-th capacitor. Since the capacitors are connected in series, the charge on each capacitor is the same: \(Q_1 = Q_2 = Q_3 = Q\)
03

Part A - Calculate the total energy stored in the capacitors in series configuration

To calculate the total energy stored in the capacitors, we will use the formula: \(U_i = \frac{1}{2}C_iV_i^2\) We will find the energy stored in each capacitor and then sum these energies to find the total energy stored in the series configuration.
04

Part B - Calculate the equivalent capacitance in parallel configuration

To find the equivalent capacitance when capacitors are connected in parallel, we can use the following formula: \(C_{eq} = C_1 + C_2 + C_3\)
05

Part B - Find the voltage across each capacitor in the parallel configuration

Since the capacitors are connected in parallel with the same potential difference, the voltage across each capacitor is the same as the potential difference: \(V_1 = V_2 = V_3 = V = 36.0 V\)
06

Part B - Calculate the total energy stored in the capacitors in parallel configuration

To calculate the total energy stored in the capacitors, we will use the same formula as in Part A: \(U_i = \frac{1}{2}C_iV_i^2\) We will find the energy stored in each capacitor and then sum these energies to find the total energy stored in the parallel configuration.

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Key Concepts

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

Series Capacitor Configuration
Capacitors connected in series create a path where each capacitor follows the next like a train in a line. In this setup, the total or equivalent capacitance (C_{eq}) of the circuit is found using the formula:
  • \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \)
Each capacitor affects the others but crucially, they all end up sharing the same charge (Q). This configuration causes the total capacitance to be less than any individual capacitor in the series. Think of it like a bottleneck; the more capacitors you have in line, the less capacity the whole line can handle. This is because the same amount of charge is spread over more capacitors, reducing the charge each one holds.
In practical terms, it's especially useful when working with high voltage applications where the voltage is distributed across several capacitors, protecting each one from high voltage that could damage them.
Parallel Capacitor Configuration
In a parallel capacitor setup, all the capacitors are connected side by side. Imagine laying three roads next to each other, instead of one long road made up of three parts. Each capacitor in a parallel circuit has the same voltage across it, simplifying calculations considerably.
  • The formula for finding the equivalent capacitance (C_{eq}) is nice and straightforward: \( C_{eq} = C_1 + C_2 + C_3 \)
Here, the total capacitance is simply the sum of all individual capacitances. So, unlike series configurations, adding more capacitors increases the total capacitance. This is because they all contribute to the capacity by bearing the same voltage. It's like having multiple tanks storing water: each tank adds its capacity to store water.
Parallel arrangements are widely used when you need a larger storage of energy without increasing the voltage each capacitor must handle.
Energy Stored in Capacitors
Capacitors store energy in the electric field created between their plates. This stored energy is often used in various applications, providing bursts of power when needed. The energy (U) stored in a capacitor can be calculated with the formula:
  • \( U = \frac{1}{2} C V^2 \)
Where C is the capacitance and V is the voltage across the capacitor.
When capacitors are in a series configuration, the amount of energy stored makes the system efficient by spreading the charge over several capacitors. However, because the equivalent capacitance is reduced, the total energy stored is also somewhat less.
In parallel configurations, the situation flips; more charge is held due to increased capacitance. This ensures higher total energy storage. Capacitors in parallel can host more energy, useful when energy density is critical.
Regardless of configuration, capacitors are key in smoothing out power supply fluctuations, storing energy temporarily, and supplying power quickly when needed, making them indispensable components in electronic circuits.

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Most popular questions from this chapter

A parallel-plate capacitor is made of two square plates \(25 \mathrm{cm}\) on a side and \(1.0 \mathrm{mm}\) apart. The capacitor is connected to a \(50.0-\mathrm{V}\) battery. With the battery still connected, the plates are pulled apart to a separation of 2.00 mm. What are the energies stored in the capacitor before and after the plates are pulled farther apart? Why does the energy decrease even though work is done in separating the plates?

What voltage must be applied to an 8.00-nF capacitor to store \(0.160 \mathrm{mC}\) of charge?

Suppose that the capacitance of a variable capacitor can be manually changed from \(100 \mathrm{pF}\) to \(800 \mathrm{pF}\) by turning a dial, connected to one set of plates by a shaft, from \(0^{\circ}\) to \(180^{\circ}\) With the dial set at \(180^{\circ}\) (corresponding to \(C=800 \mathrm{pF}\) ), the capacitor is connected to a \(500-\mathrm{V}\) source. After charging, the capacitor is disconnected from the source, and the dial is turned to \(0^{\circ} .\) If friction is negligible, how much work is required to turn the dial from \(180^{\circ}\) to \(0^{\circ} ?\)

A parallel-plate capacitor with capacitance \(5.0 \mu \mathrm{F}\) is charged with a 12.0 -V battery, after which the battery is disconnected. Determine the minimum work required to increase the separation between the plates by a factor of 3.

A parallel-plate capacitor has charge of magnitude \(9.00 \mu \mathrm{F}\) on each plate and capacitance \(3.00 \mu \mathrm{C}\) when there is air between the plates. The plates are separated by \(2.00 \mathrm{mm} .\) With the charge on the plates kept constant, a dielectric with \(\kappa=5\) is inserted between the plates, completely filling the volume between the plates. (a) What is the potential difference between the plates of the capacitor, before and after the dielectric has been inserted? (b) What is the electrical field at the point midway between the plates before and after the dielectric is inserted?

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