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Chapter 18: Problem 115

A 2.00 - \(\mu\) F capacitor is charged using a \(5.00-\mathrm{V}\) battery and a 3.00 - \(\mu\) F capacitor is charged using a \(10.0-\mathrm{V}\) battery. (a) What is the total energy stored in the two capacitors? (b) The batteries are disconnected and the two capacitors are connected together \((+\text { to }+\) and \- to \(-\) ). Find the charge on each capacitor and the total energy in the two capacitors after they are connected. (c) Explain what happened to the "missing" energy. [Hint: The wires that connect the two have some resistance.]

Short Answer

Expert verified
Question: Calculate the following for the given capacitors: (a) the total initial energy stored, (b) the charge and total energy stored after connecting them, and (c) explain the "missing" energy. Capacitor 1: capacitance = 2.00 μF, voltage = 5 V Capacitor 2: capacitance = 3.00 μF, voltage = 10 V a) Total initial energy stored: ______ b) Charge on each capacitor after connecting them: Capacitor 1: ______ Capacitor 2: ______ Total energy stored after connecting: ______ c) Explanation for the "missing" energy: __________________________________________________

Step by step solution

01

Find the initial energy stored in both capacitors#

Using the capacitor energy formula \(E = \frac{1}{2}CV^2\), we can find the initial energy in both capacitors: For the 2 μF capacitor (\(C_1\)): \(E_1 = \frac{1}{2}(2\times 10^{-6}\,F)(5\,V)^2 = 25\times 10^{-6}\,J\) For the 3 μF capacitor (\(C_2\)): \(E_2 = \frac{1}{2}(3\times 10^{-6}\,F)(10\,V)^2 = 150\times 10^{-6}\,J\)
02

Calculate the total initial energy stored#

Now, add the initial energies of both capacitors: \(E_\text{total} = E_1 + E_2 = 25\times 10^{-6}\,J + 150\times 10^{-6}\,J = 175\times 10^{-6}\,J\)
03

Calculate the initial charge on each capacitor#

Calculate the initial charge on each capacitor using the formula \(Q = CV\). For the 2 μF capacitor (\(C_1\)): \(Q_1 = C_1V_1 = (2\times10^{-6}\,F)(5\,V) = 10\times 10^{-6}\,C\) For the 3 μF capacitor (\(C_2\)): \(Q_2 = C_2V_2 = (3\times 10^{-6}\,F)(10\,V) = 30\times 10^{-6}\,C\)
04

Calculate the common voltage and the charge on each capacitor after connecting them#

After connecting the capacitors, the total charge is conserved, and they will share a common voltage. Total charge: \(Q_\text{total} = Q_1 + Q_2 = 10\times10^{-6}\,C + 30\times10^{-6}\,C = 40\times10^{-6}\,C\) The common voltage: \(V_\text{common} = \frac{Q_\text{total}}{C_1 + C_2} = \frac{40\times 10^{-6}\,C}{(2 + 3)\times 10^{-6}\,F} = 8\,V\) Now we can calculate the charge on each capacitor after connection: For the 2 μF capacitor (\(C_1\)): \(Q_1' = C_1V_\text{common} = (2\times10^{-6}\,F)(8\,V) = 16\times 10^{-6}\,C\) For the 3 μF capacitor (\(C_2\)): \(Q_2' = C_2V_\text{common} = (3\times 10^{-6}\,F)(8\,V) = 24\times 10^{-6}\,C\)
05

Calculate the total energy stored after connecting the capacitors#

Now, calculate the energy stored in each capacitor using the formula \(E = \frac{1}{2}CV^2\) after connection: For the 2 μF capacitor (\(C_1\)): \(E_1' = \frac{1}{2}(2\times 10^{-6}\,F)(8\,V)^2 = 64\times 10^{-6}\,J\) For the 3 μF capacitor (\(C_2\)): \(E_2' = \frac{1}{2}(3\times 10^{-6}\,F)(8\,V)^2 = 96\times 10^{-6}\,J\) Total energy after connecting the capacitors: \(E_\text{total}' = E_1' + E_2' = 64\times 10^{-6}\,J + 96\times 10^{-6}\,J = 160\times 10^{-6}\,J\)
06

Explain the "missing" energy#

Before connecting the capacitors, the total energy stored was \(175\times 10^{-6}\,J\). After connecting the capacitors, the total energy stored is \(160\times 10^{-6}\,J\). The difference between these two values is \(15\times 10^{-6}\,J\). This "missing" energy can be explained by the fact that there is some resistance encountered by the current in the wires while connecting the capacitors. This resistance leads to energy loss in the form of heat.

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