91Ó°ÊÓ

The initial energy stored in a capacitor can be calculated using the formula \(E = \frac{1}{2}CV^2\), where \(E\) is the energy, \(C\) is the capacitance, and \(V\) is the initial voltage across the capacitor. In this problem, we are given the capacitance \(C = 90.0\,\mu \mathrm{F}\), but the initial voltage \(V\) is not given. However, we do not need to calculate the initial voltage, as we only need to find the time it takes for the energy to decrease by \(80\%\). The initial voltage will cancel out when we find the ratio of the energy at a certain time compared to the initial energy.

The voltage across a capacitor during discharging decreases exponentially over time and can be expressed by the formula \(V(t) = V_0e^{-t/RC}\), where \(V(t)\) is the voltage across the capacitor at time \(t\), \(V_0\) is the initial voltage across the capacitor, \(R\) is the resistance of the resistor, and \(C\) is the capacitance of the capacitor. For this problem, we know that the resistance of the resistor is \(60.0\,\Omega\), and the capacitance of the capacitor is \(90.0\,\mu \mathrm{F}\). Thus, we can rewrite the formula for the voltage across the capacitor as \(V(t) = V_0e^{-t/(60.0 \cdot 90.0\,\mu)}\).

To find the energy at a certain time, we first need to find the voltage across the capacitor at that time using the formula we derived in Step 2. After finding the voltage, we can calculate the energy using the formula \(E(t) = \frac{1}{2}CV(t)^2\).

We are asked to find the time at which the energy falls to \(20\%\) of its initial energy, which corresponds to an \(80\%\) loss of energy. Thus, we need to find the time \(t\) at which \(E(t) = 0.2E_0\), where \(E_0\) is the initial energy. Using the formula for the energy at a certain time, we can rewrite this equation as \(\frac{1}{2}C(V_0^2 - V(t)^2) = 0.2E_0\), and substitute the formula for the voltage across the capacitor during discharging, giving us \(\frac{1}{2}C(V_0^2 - V_0^2e^{-2t/RC}) = 0.2E_0\). Now, we can divide both sides of the equation by \(E_0\), which gives us \(1 - e^{-2t/RC} = 0.2\), and solve for \(t\). \(e^{-2t/RC} = 0.8\). Taking the natural logarithm of both sides gives us \(-2t/RC = \ln(0.8)\). Finally, we can solve for \(t\): \(t = -\frac{RC}{2}\ln(0.8) = -\frac{60.0 \cdot 90.0\,\mu}{2}\ln(0.8) \approx 26.29\, \mathrm{ms}\). So, it will take approximately \(26.29\,\mathrm{ms}\) for the capacitor to lose \(80\%\) of its initial energy.