A \(0.30-\mathrm{H}\) inductor and a \(200.0-\Omega\) resistor are connected in series to a \(9.0-\mathrm{V}\) battery. (a) What is the maximum current that flows in the circuit? (b) How long after connecting the battery does the current reach half its maximum value? (c) When the current is half its maximum value, find the energy stored in the inductor, the rate at which energy is being stored in the inductor, and the rate at which energy is dissipated in the resistor. (d) Redo parts (a) and (b) if, instead of being negligibly small, the internal resistances of the inductor and battery are \(75 \Omega\) and \(20.0 \Omega\) respectively.

: First, we can use Kirchhoff's loop rule to find an equation that governs the circuit: \(V_{battery} - V_{inductor} - V_{resistor} = 0\) We know that \(V_{inductor} = L\frac{dI}{dt}\) and \(V_{resistor} = IR\). So, \(V_{battery} - L\frac{dI}{dt} - IR = 0\) Now, let's find the maximum current, \(I_{max}\) by setting \(\frac{dI}{dt} = 0\). We have: \(V_{battery} - IR_{max} = 0\) Solving for \(I_{max}\): \(I_{max} = \frac{V_{battery}}{R} = \frac{9.0\mathrm{V}}{200.0\Omega} = 0.045\mathrm{A}\)

: For this part, we want to find the time \(t\) when \(I = \frac{I_{max}}{2}\). Let's use the equation from part (a): \(V_{battery} - L\frac{dI}{dt} - IR = 0\) Now let's rearrange the equation to solve for \(\frac{dI}{dt}\): \(\frac{dI}{dt} = \frac{V_{battery} - IR}{L}\) Integrating both sides, we have: \(\int_{0}^{I_\frac{max}{2}}\frac{dI}{RI} = \int_{0}^{t}\frac{V_{battery}(dt)}{L}\) Solving the integrals: \(-\frac{R}{L} \ln (1 - \frac{I_\frac{max}{2}}{I_{max}}) = \frac{V_{battery}t}{L}\) Since \(0 < I_\frac{max}{2} < I_{max}\), we can solve for \(t\): \(t = -\frac{L}{V_{battery}}\ln (1 - \frac{1}{2}) = \frac{0.30\mathrm{H}}{9.0\mathrm{V}}\ln(2) \approx 0.0232\mathrm{s}\)

: Now, let's find the energy stored in the inductor, the rate at which energy is being stored in the inductor, and the rate at which energy is dissipated in the resistor. - Energy stored in the inductor: \(U_{inductor} = \frac{1}{2}LI^2 = \frac{1}{2}(0.30\mathrm{H})\left(\frac{I_{max}}{2}\right)^2 = \frac{1}{8}(0.30\mathrm{H})(0.045\mathrm{A})^2 \approx 3.04\times10^{-4}\mathrm{J}\) - The rate at which energy is being stored in the inductor: \(\frac{dU_{inductor}}{dt} = L\frac{d}{dt}\left(\frac{1}{2}I^2\right) = LI\frac{dI}{dt} = V_{inductor}\) So, \(\frac{dU_{inductor}}{dt}= V_{battery} - V_{resistor} = 9.0\mathrm{V} - (200.0\Omega)(0.0225\mathrm{A}) \approx 4.5\mathrm{V}\) - The rate at which energy is dissipated in the resistor: \(\frac{dU_{resistor}}{dt} = I^2R = (0.0225\mathrm{A})^2(200.0\Omega) \approx 0.101\mathrm{W}\)

: Now we have to redo parts (a) and (b) considering that the internal resistances of the inductor and battery are \(75\Omega\) and \(20.0\Omega\), respectively. Our new total resistance is: \(R_{total} = R + R_{inductor} + R_{battery} = 200 + 75 + 20.0\Omega = 295\Omega\) - Maximum current in the circuit \(I_{max}' = \frac{V_{battery}}{R_{total}} = \frac{9.0\mathrm{V}}{295\Omega} \approx 0.0305\mathrm{A}\) - Time to reach half the maximum current: Using the same procedure as before but replacing \(R\) with \(R_{total}\), we have \(t' = \frac{0.30\mathrm{H}}{9.0\mathrm{V}}\ln(2) \times \left(\frac{200\Omega}{295\Omega}\right) \approx 0.0156\mathrm{s}\) So, the maximum current in the circuit considering the internal resistances is \(0.0305\mathrm{A}\), and the time to reach half the maximum current is \(0.0156\mathrm{s}\).