Chapter 30 - Electromagnetic Induction - Exercises and Problems - Page 874: 77

a.) $\displaystyle I_0 = \frac{\Delta V_{bat}}{R}$ b.) $I(t) = \displaystyle I_0(1 - e^{-Rt/L})$ c.) See graph

a.) When the switch has been open for a long time and that there is a steady current in the circuit such that $\displaystyle \frac{dI}{dt} = 0$. Doing so and using Kirchoff's loop law, we find that $\Delta V_{bat} - IR = 0$ $\displaystyle \frac{\Delta V_{bat}}{R} = I_0$ The question should no call this $I_0$ though, because $I_0$ gives the impression that this is the initial current in the circuit when the switch is closed. THIS IS NOT THE INITIAL CURRENT. This is the current when the switch has been closed for a long time, it is the maximum current. It should be called $I_{max}$ It would, however, if the switch was closed for a long time such that there was a steady current and then the switch was suddenly opened. Then this would be the initial current. --- b.) After the switch has been closed, Kirchoff's loop law gives us $\Delta V_{bat} - IR - L\displaystyle \frac{dI}{dt} = 0$ We now have a first-order linear differential equation. This can be solved with separation of variables. $\displaystyle L\frac{dI}{dt} = \Delta V_{bat} - IR$ $\displaystyle \int\frac{dI}{\Delta V_{bat} - IR} = \int\frac{dt}{L}\qquad\qquad\qquad$ let $\quad u = \Delta V_{bat} - IR$ $\qquad\qquad\qquad\qquad\qquad\qquad\quad$such that $\qquad du= -R \cdot dI\qquad\rightarrow\qquad\displaystyle -\frac{1}{R}du = dI$ $\displaystyle \int\frac{-\frac{1}{R}du}{u} = \int\frac{dt}{L}$ $\displaystyle -\frac{1}{R}\int\frac{du}{u} = \int\frac{dt}{L}$ $\displaystyle -\frac{1}{R}\ln|u| = \frac{t}{L} + C$ $\displaystyle -\frac{1}{R}\ln|\Delta V_{bat} - IR| = \frac{t}{L} +C $ $\displaystyle \ln|\Delta V_{bat} - IR| = -\frac{Rt}{L} + C\qquad\qquad$ the reason why it's not $\displaystyle -\frac{Rt}{L} -RC$ is because $C$, the $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ constant of integration, hasn't taken on an initial condition yet. $\displaystyle \ln|\Delta V_{bat} - IR| = -\frac{Rt}{L} + C$ $\displaystyle e^{\ln|\Delta V_{bat} - IR|} = e^{-\frac{Rt}{L} + C}$ $\displaystyle \Delta V_{bat} - IR = e^{-Rt/L}(e^C)$ $IR = \Delta V_{bat} - Ce^{-Rt/L}$ ----------------------------------- $I(0) = 0 \ne I_0$ $(0)R = \Delta V_{bat} - Ce^0$ $0 = \Delta V_{bat} - C(1)$ $\Delta V_{bat} = C$ ----------------------------------- $IR = \Delta V_{bat} - \Delta V_{bat}e^{-Rt/L}$ $I = I(t) = \displaystyle \frac{\Delta V_{bat}}{R}(1 - e^{-Rt/L}) = I_0(1 - e^{-Rt/L})$ c.) The graph is shown below: