A coil has an inductance of \(0.15 \mathrm{H}\) and a resistance of 33 \(\Omega\). The coil is connected to a 6.0 - \(V\) ideal battery. When the current reaches half its maximum value: (a) At what rate is magnetic energy being stored in the inductor? (b) At what rate is energy being dissipated? (c) What is the total power that the battery supplies?

Knowing the resistance R and voltage V, we can find the maximum current (I_max) using Ohm's law: \(I_{max} = \frac{V}{R}\) Plug in our given values of R and V: \(I_{max} = \frac{6.0\,\text{V}}{33\,\Omega} = 0.1818\,\mathrm{A}\) When the current reaches half its maximum value: \(I = \frac{I_{max}}{2} = 0.1818 \times 0.5 = 0.0909 \,\mathrm{A}\)

The rate of change of energy stored in the inductor is given by the formula: \(\frac{dW_L}{dt} = L I \frac{dI}{dt}\) Here we need to find \(\frac{dI}{dt}\). Using Kirchhoff's voltage law and knowing that the voltage across the inductor is \(V_L = L\frac{dI}{dt}\) and the voltage across the resistor is \(V_R = IR\), we can write: \(V = V_R + V_L = IR + L\frac{dI}{dt}\) Now we need to find \(\frac{dI}{dt}\) from the equation above: \(\frac{dI}{dt} = \frac{V - IR}{L}\) Plugging the values for I, V and L, \(\frac{dI}{dt} = \frac{6.0\,\text{V} - 33\,\Omega \times 0.0909\,\mathrm{A}}{0.15\,\mathrm{H}} = 40\,\mathrm{A/s}\) Now we have all values needed to find \(\frac{dW_L}{dt}\): \(\frac{dW_L}{dt} = 0.15\,\mathrm{H} \times 0.0909\,\mathrm{A} \times 40\,\mathrm{A/s} = 0.5454\,\mathrm{W}\) Thus, the rate at which magnetic energy is being stored in the inductor when the current reaches half its maximum value is \(\boxed{0.5454\,\mathrm{W}}\).

The rate of energy dissipated in the resistor is given by the formula: \(P_R = I^2R\) Plugging the values for I and R, \(P_R = (0.0909\,\mathrm{A})^2 \times 33\,\Omega = 0.2727\,\mathrm{W}\) Hence, the rate at which energy is being dissipated when the current reaches half its maximum value is \(\boxed{0.2727\,\mathrm{W}}\).

The total power supplied by the battery is the sum of the rate of energy storage in the inductor and the rate of energy dissipation in the resistor, \(P_{total} = \frac{dW_L}{dt} + P_R = 0.5454\,\mathrm{W} + 0.2727\,\mathrm{W} = \boxed{0.8181\,\mathrm{W}}\) Thus, the total power supplied by the battery when the current reaches half its maximum value is \(0.8181\,\mathrm{W}\).