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Unformatted text preview: Now, i = dq/dt o dq/dt = ( /R)e- t/RC o dq = ( /R)e- t/RC dt o I (integral from q to 0) dq = ( /R) I (integral from t to 0) e- t/RC Remember that I (integral) e-ax dx = -(1/a) e-ax o Therefore, q = ( /R)[-RC][e- t/RC ] (evaluated from t to 0) o q = - C [ e- t/RC-1] Or similarly. .. q(t) = C [ 1 -e- t/RC ] q(t) = C [ 1 -e- t/ ], where is the time constant which equals RC. See Figure 28-14 on page 686 on the text for graphs the above information. Discharging a Capacitor At t > 0, Kirchoff's rule gives, (q/C) - iR =0. o But since i= dq/dt. .... (q/C) + R(dq/dt) =0 o Or. ... R(dq/dt) = -(q/C) o 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 o q(t) = q max e-t/RC d/dt [q(t) = q max e-t/RC ] o-i = dq/dt = q max (1/RC)(- e-t/RC ) o So, i = ( q max /RC)(- e-t/RC ) o i = i o e-t/RC See problem 72 from Chapter 28 of the text...
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