Question

A 12.0-V battery is attached to a 2.00 -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

**Short Answer:** The energy stored in the capacitor when it is fully charged is 144 mJ. The energy dissipated by the resistor as the capacitor is charging is also 144 mJ. This is in accordance with the conservation of energy, as the total energy supplied by the battery is divided between the capacitor and the resistor.

Step-by-step solution

Find the energy stored in the capacitor

Use the formula for the energy stored in a capacitor: \(U_c=\frac{1}{2}C V^2\), where \(U_c\) is the energy stored in the capacitor, \(C\) is the capacitance, and \(V\) is the voltage across the capacitor. In this case, \(C=2.00\,\text{mF}=2.00\times10^{-3}\,\text{F}\) and \(V=12.0\,\text{V}\). Now plug in the values: \(U_c=\frac{1}{2}(2.00\times10^{-3}\,\text{F})(12.0\,\text{V})^2\).

Compute the energy stored in the capacitor

Compute the formula from the previous step: \(U_c = \frac{1}{2}(2.00\times10^{-3}\,\text{F})(144\,\text{V}^2) = 144\,\text{mJ}\). Thus, the energy stored in the capacitor when it is fully charged is 144 mJ.

Calculate the energy dissipated by the resistor using conservation of energy

The energy provided by the battery is divided between the capacitor and the resistor, so the energy dissipated by the resistor is equal to the total energy provided by the battery minus the energy stored in the capacitor. First, let's find the total energy supplied by the battery by using the formula: \(U_b=QV\), where \(U_b\) is the energy supplied by the battery, \(Q\) is the charge stored in the capacitor, and \(V\) is the voltage across the capacitor. We know that \(Q = CV\), so we can rewrite the formula as: \(U_b = (CV)V = CV^2\). Plug in the values for \(C\) and \(V\): \(U_b = (2.00\times10^{-3}\,\text{F})(12.0\,\text{V})^2 = 288\,\text{mJ}\). This is the total energy supplied by the battery. Now subtract the energy stored in the capacitor: \(U_r = U_b - U_c = 288\,\text{mJ} - 144\,\text{mJ} = 144\,\text{mJ}\).

Interpret the results

The energy stored in the capacitor when it is fully charged is equal to 144 mJ, and the energy dissipated by the resistor as the capacitor is charging is also 144 mJ. Conservation of energy holds, as the total energy supplied by the battery, 288 mJ, was divided between the capacitor and the resistor.