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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 capacitor \(C\) is initially uncharged. At time \(t=0,\) the capacitor is attached through a resistor \(R\) to a battery. The energy stored in the capacitor increases, eventually reaching a value \(U\) as \(t \rightarrow \infty\), After a time equal to the time constant \(\tau=R C\), the energy stored in the capacitor is given by a) \(U / e\). c) \(U(1-1 / e)^{2}\) b) \(U / e^{2}\) d) \(U(1-1 / e)\).

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?

The ammeter your physics instructor uses for in-class demonstrations has internal resistance \(R_{\mathrm{i}}=75 \Omega\) and measures a maximum current of \(1.5 \mathrm{~mA}\). The same ammeter can be used to measure currents of much greater magnitudes by wiring a shunt resistor of relatively small resistance, \(R_{\text {shunt }}\), in parallel with the ammeter. (a) Sketch the circuit diagram, and explain why the shunt resistor connected in parallel with the ammeter allows it to measure larger currents. (b) Calculate the resistance the shunt resistor has to have to allow the ammeter to measure a maximum current of 15 A.

You want to make an ohmmeter to measure the resistance of unknown resistors. You have a battery with voltage \(\mathrm{V}_{\mathrm{emf}}=9.00 \mathrm{~V}\), a variable resistor, \(R,\) and an ammeter that measures current on a linear scale from 0 to \(10.0 \mathrm{~mA}\) a) What resistance should the variable resistor have so that the ammeter gives its full-scale (maximum) reading when the ohmmeter is shorted? b) Using the resistance from part (a), what is the unknown resistance if the ammeter reads \(\frac{1}{4}\) of its full scale?

Many electronics devices can be dangerous even after they are shut off. Consider an RC circuit with a \(150 .-\mu \mathrm{F}\) capacitor and a \(1.00-\mathrm{M} \Omega\) resistor connected to a \(200 .-\mathrm{V}\) power source for a long time and then disconnected and shorted, as shown in the figure. How long will it be until the potential difference across the capacitor drops to below \(50.0 \mathrm{~V} ?\)
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