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Chapter 26: Problem 48

A \(12.0-V\) battery is attached to a \(2.00-\mathrm{mF}\) capacitor and a \(100 .-\Omega\) resistor. Once the capacitor is fully charged, what is the energy stored in it? What is the energy dissipated as heat by the resistor as the capacitor is charging?

Short Answer

Expert verified
Answer: The energy stored in the capacitor when it is fully charged is 0.144 J, and the energy dissipated as heat by the resistor as the capacitor is charging is also 0.144 J.

Step by step solution

01

Calculate the energy stored in the fully charged capacitor

We are given that the voltage (\(V=12.0 V\)) and capacitance (\(C=2.00 mF\)). Using the formula for the energy stored in a capacitor \(W_c = \frac{1}{2}CV^2\), we can calculate the energy stored: \(W_c = \frac{1}{2}(2.00 \times 10^{-3} F)(12.0 V)^2\) \(W_c = 0.144 J\) So the energy stored in the capacitor when it is fully charged is \(0.144 J\).
02

Calculate the total energy supplied by the battery

To calculate the energy supplied by the battery, we can use the formula \(W_b = QV\), where \(Q\) is the charge transferred. Since the charge on the capacitor is given by \(Q = CV\), we can rewrite the formula as \(W_b = CV^2\). Now, substituting the given values: \(W_b = (2.00 \times 10^{-3} F)(12.0 V)^2\) \(W_b = 0.288 J\) The total energy supplied by the battery is \(0.288 J\).
03

Find the energy dissipated as heat by the resistor

Now that we have the energy supplied by the battery (\(W_b = 0.288 J\)) and the energy stored in the capacitor (\(W_c = 0.144 J\)), we can find the energy dissipated as heat by the resistor using the conservation of energy principle: \(W_b = W_c + W_r\). Rearranging the equation to solve for \(W_r\): \(W_r = W_b - W_c\) \(W_r = 0.288 J - 0.144 J\) \(W_r = 0.144 J\) The energy dissipated as heat by the resistor as the capacitor is charging is \(0.144 J\).

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Most popular questions from this chapter

A parallel plate capacitor with \(C=0.050 \mu \mathrm{F}\) has a separation between its plates of \(d=50.0 \mu \mathrm{m} .\) The dielectric that fills the space between the plates has dielectric constant \(\kappa=2.5\) and resistivity \(\rho=4.0 \cdot 10^{12} \Omega \mathrm{m} .\) What is the time constant for this capacitor? (Hint: First calculate the area of the plates for the given \(C\) and \(\kappa\), and then determine the resistance of the dielectric between the plates.)

A circuit consists of a source of emf, a resistor, and a capacitor, all connected in series. The capacitor is fully charged. How much current is flowing through it? a) \(i=V / R\) b) zero c) neither (a) nor (b)

An RC circuit has a time constant of 3.1 s. At \(t=0\), the process of charging the capacitor begins. At what time will the energy stored in the capacitor reach half of its maximum value?

If the capacitor in an \(\mathrm{RC}\) circuit is replaced with two identical capacitors connected in series, what happens to the time constant for the circuit?

Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=\) 0 , the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega\), that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q\) ?
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