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Chapter 24: Problem 48

(II) A \(2.20-\mu \mathrm{F}\) capacitor is charged by a \(12.0 . \mathrm{V}\) battery. It is disconnected from the battery and then connected to an uncharged \(3.50-\mu \mathrm{F}\) capacitor (Fig. \(20 ) .\) Determine the total stored energy \((a)\) before the two capacitors are connected, and (b) after they are connected. (c) What is the change in energy?

Short Answer

Expert verified
(a) Initial energy is \(1.584 \times 10^{-3}\) J. (b) Final energy is \(6.11 \times 10^{-5}\) J. (c) Energy change is \(1.5239 \times 10^{-3}\) J.

Step by step solution

01

Calculate Initial Stored Energy in Single Capacitor

Initially, the charged capacitor has a capacitance of \(2.20 \mu F\) and a voltage of \(12.0 V\). The formula for energy stored in a capacitor is \[ E = \frac{1}{2} C V^2 \]Substituting the given values, \[ E = \frac{1}{2} \times 2.20 \times 10^{-6} F \times (12.0 V)^2 \]\[ E = \frac{1}{2} \times 2.20 \times 10^{-6} F \times 144 \text{ V}^2 \]\[ E = 1.584 \times 10^{-3} J \]Thus, the initial stored energy is \(1.584 \times 10^{-3}\) joules.
02

Calculate Final Voltage After Connecting Capacitors

When the capacitors are connected, charge is conserved. The total capacitance when they are in parallel is \[ C_{total} = 2.20 \mu F + 3.50 \mu F = 5.70 \mu F \] The initial charge \(Q\) on the charged capacitor is \[ Q = C \times V = 2.20 \mu F \times 12.0 V = 26.4 \mu C \]After connection, the voltage \(V_f\) across both capacitors is given by \[ V_f = \frac{Q}{C_{total}} = \frac{26.4 \mu C}{5.70 \mu F} \approx 4.63 V \]
03

Calculate Total Stored Energy After Connection

The total energy stored in the connected capacitors can be calculated as:\[ E_f = \frac{1}{2} C_{total} V_f^2 \]Substitute the previously calculated values:\[ E_f = \frac{1}{2} \times 5.70 \times 10^{-6} F \times (4.63 V)^2 \]\[ E_f = \frac{1}{2} \times 5.70 \times 10^{-6} F \times 21.4469 \text{ V}^2 \]\[ E_f = 6.11 \times 10^{-5} J \]
04

Calculate the Change in Energy

The change in energy is the initial energy minus the final energy:\[ \Delta E = E_{initial} - E_f \]\[ \Delta E = 1.584 \times 10^{-3} J - 6.11 \times 10^{-5} J \]\[ \Delta E = 1.5239 \times 10^{-3} J \]Thus, the energy change is \(1.5239 \times 10^{-3}\) joules.

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Key Concepts

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

Conservation of Charge
When discussing capacitors, especially in a circuit like the one in the exercise, conservation of charge is a crucial concept. Charge conservation states that the total electric charge in an isolated system remains constant over time, regardless of other changes that take place within the system.

In the context of capacitors, when two capacitors are connected in parallel, the total charge before the connection equals the total charge after the connection. The formula to express this is:
  • Before connection: \( Q_{initial} = C_1 imes V_1 \)
  • After connection: \( Q_{total} = C_{total} imes V_{final} \)
By setting these equal, we get: \( Q_{initial} = Q_{total} \) This allows us to calculate the resultant voltage across the capacitors once connected, helping in further computations like energy calculations.
Parallel Capacitors
In a circuit when capacitors are connected together, they can be arranged in series or parallel. In our case, they are connected in parallel.

When capacitors are in parallel, their total capacitance is the sum of the individual capacitances. This is much simpler than dealing with series capacitors, where the calculation is a bit more complex.
  • The formula for the total capacitance in parallel: \( C_{total} = C_1 + C_2 \)
Knowing this helps calculate the newly shared voltage after connection, essential for finding out the new energy stored in the system. This approach changes the initial energy distribution because both capacitors share the initial charge resulting in a new, combined capacitance.
Stored Energy Calculation
Calculating stored energy in capacitors involves using the fundamental formula: \( E = \frac{1}{2} C V^2 \).

This formula represents the energy stored due to the electric field in a charged capacitor, where\( C \) is the capacitance and \( V \) is the voltage.
  • Initial energy is calculated using the capacitance and voltage before connecting the capacitors in parallel.
  • After connecting, the capacitors now share a total capacitance and a newly calculated voltage. The stored energy is recalculated using these new values.
The comparison between the initial and final energies shows an apparent decrease, signifying some of the initial energy is lost. This is typical in such problems, usually due to small losses in the connecting wires or potential other forms of dissipation. Understanding this helps predict the expected change and provides deeper insight into real-life electrical systems.

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Most popular questions from this chapter

Consider three capacitors, of capacitance \(3600 \mathrm{pF}\) \(5800 \mathrm{pF},\) and \(0.0100 \mu \mathrm{F}\). What maximum and minimum capacitance can you form from these? How do you make the connection in each case?

The electric field between the plates of a paper-separated \((K=3.75)\) capacitor is \(9.21 \times 10^{4} \mathrm{V} / \mathrm{m} .\) The plates are 1.95 \(\mathrm{mm}\) apart and the charge on each plate is 0.675\(\mu \mathrm{C}\) . Determine the capacitance of this capacitor and the area of each plate.

A high-voltage supply can be constructed from a variable capacitor with interleaving plates which can be rotated as in Fig. \(24-36 .\) A version of this type of capacitor with more plates has a capacitance which can be varied from \(10 \mathrm{pF}\) to \(1 \mathrm{pF}\). (a) Initially, this capacitor is charged by a \(7500-\mathrm{V}\) power supply when the capacitance is \(8.0 \mathrm{pF}\). It is then disconnected from the power supply and the capacitance reduced to \(1.0 \mathrm{pF}\) by rotating the plates. What is the voltage across the capacitor now? (b) What is a major disadvantage of this as a high-voltage power supply?

A parallel-plate capacitor with plate area \(2.0 \mathrm{~cm}^{2}\) and airgap separation \(0.50 \mathrm{~mm}\) is connected to a \(12-\mathrm{V}\) battery, and fully charged. The battery is then disconnected. ( \(a\) ) What is the charge on the capacitor? (b) The plates are now pulled to a separation of \(0.75 \mathrm{~mm}\). What is the charge on the capacitor now? (c) What is the potential difference across the plates now? ( \(d\) ) How much work was required to pull the plates to their new separation?

(1I) Given three capacitors, \(C_{1}=2.0 \mu \mathrm{F}, C_{2}=1.5 \mu \mathrm{F},\) and \(C_{3}=3.0 \mu \mathrm{F}, \quad\) what arrangement of parallel and series connections with a \(12-\mathrm{V}\) battery will give the minimum voltage drop across the \(2.0-\mu \mathrm{F}\) capacitor? What is the minimum voltage drop?
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