Tuesday, March 16, 1999
Announcements: Exam Thursday!
Material Covered in today's lecture WILL NOT be on midterm 2

Lecture notes:

• At time t = 0, the switch is closed.
• At some later time, we can use Kirchhoff's rule,   x - iR - (q/C) = 0   ( equation A)
• Note that at t = 0, q = 0. Therefore,  x - i_oR = 0 .
• So,  i_o = x/R . This is the current at t = 0.
• For t > 0, i decreases from  i_o.
• Also note that as t approaches infinity, the capacitor will become fully charged and the current, i , will stop flowing. i = 0.
• In Equation A, for very large t, x - (q_max/C) = 0. Therefore,  q_max= Cx.
• To determine the time dependence of i and q, (d/dt) [  x - iR - (q/C) ]
• = 0 - R(di/dt) - (1/C)(dq/dt)
• But, i =   dq/dt
• Therefore,  R(di/dt) + (1/C)i =0
• R(di/dt) = - i/C  or   di/i = - (1/RC) dt
• I(integral from i to i_o) di/i = -(1/RC) I(integral from t to 0) dt
• ln (i/i_o) = - t/ RC
• i(t) = i_o e^{- t/RC}    where e is the exponential not an electron
• Therefore, i (t) = (x/R)e^{- t/RC}
• Current drops off exponentially
• Now, i = dq/dt
• dq/dt = (x/R)e^{- t/RC}
• dq = (x/R)e^{- t/RC}  dt
• I (integral from q to 0) dq = (x/R) I (integral from t to 0) e^{- t/RC}
• Remember that I (integral) e^-ax dx = -(1/a) e^-ax
• Therefore, q = (x/R)[-RC][e^{- t/RC} ] (evaluated from t to 0)
• q = -xC [ e^{- t/RC} -1]
• Or similarly... q(t) = Cx [ 1 - e^{- t/RC} ]
• q(t) = Cx [ 1 - e^{- t/t} ], where t is the time constant which equals RC.
• See Figure 28-14 on page 686 on the text for graphs the above information.

• At t > 0, Kirchoff's rule gives, (q/C) - iR =0.
• But since i= dq/dt..... (q/C) + R(dq/dt) =0
• Or.... R(dq/dt) = -(q/C)
• Or.... dq/q = - (1/RC) dt
• So, I( integral from q to q_max) dq/d = -(1/RC) I (integral from t to 0) dt
• ln (q/ q_max) = -t/RC
• q(t) = q_max e^-t/RC

• d/dt [q(t) = q_max e^-t/RC ]
• -i = dq/dt = q_max (1/RC)(- e^-t/RC )
• So,     i = ( q_max /RC)(- e^-t/RC )
• i = i_o e^-t/RC

• See problem 72 from Chapter 28 of the text