A 20 - \(\mu\) F capacitor is discharged through a 5 -k\Omega resistor. The initial charge on the capacitor is \(200 \mu \mathrm{C}\) (a) Sketch a graph of the current through the resistor as a function of time. Label both axes with numbers and units. (b) What is the initial power dissipated in the resistor? (c) What is the total energy dissipated?

To sketch the graph, we will use the current discharge formula: \(I(t) = \frac{Q_0}{R}\exp\left(-\frac{t}{RC}\right)\). For our given values, \(Q_0 = 200 \times 10^{-6} C\), \(R = 5 \times 10^3 \Omega\), and \(C = 20 \times 10^{-6} F\). Plug in these values to get: \(I(t) = \frac{200 \times 10^{-6}}{5 \times 10^3}\exp\left(-\frac{t}{(5 \times 10^3)(20 \times 10^{-6})}\right)\). Now we can plot this equation, which will result in a decreasing exponential curve. The x-axis represents time (in seconds), and the y-axis represents the current (in amperes).

The initial current \(I(0)\) can be found by substituting \(t = 0\) in our current equation: \(I(0) = \frac{200 \times 10^{-6}}{5 \times 10^3}\exp\left(0\right) = \frac{200 \times 10^{-6}}{5 \times 10^3}\). Now, we can calculate the initial power dissipated using the formula \(P = I^2R\): \(P_0 = \left(\frac{200 \times 10^{-6}}{5 \times 10^3}\right)^2 (5 \times 10^3) = 8 \times 10^{-3} W\). Therefore, the initial power dissipated in the resistor is \(8 \times 10^{-3} W\).

To find the total energy dissipated, we first need to find the formula for power as a function of time, which is \(P(t) = I^2(t)R\). Substitute our current formula: \(P(t) = \left(\frac{200 \times 10^{-6}}{5 \times 10^3}\exp\left(-\frac{t}{(5 \times 10^3)(20 \times 10^{-6})}\right)\right)^2 (5 \times 10^3)\). Now, we need to integrate the power formula from \(t = 0\) to \(t = +\infty\): \(E = \int^{+\infty}_0 P(t) dt = \int^{\infty}_0 \left(\frac{200 \times 10^{-6}}{5 \times 10^3}\exp\left(-\frac{t}{(5 \times 10^3)(20 \times 10^{-6})}\right)\right)^2 (5 \times 10^3) dt\). This integration evaluates to: \(E = Q_0^2 / (2C) = \frac{(200 \times 10^{-6})^2}{2(20 \times 10^{-6})} = 2 \times 10^{-3} J\) Therefore, the total energy dissipated in the resistor is \(2 \times 10^{-3} J\).