91Ó°ÊÓ

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 \(0.8181\,\mathrm{W}\).