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Unformatted text preview: • Now, i = dq/dt o dq/dt = ( ξ /R)e t/RC o dq = ( ξ /R)e t/RC dt o I (integral from q to 0) dq = ( ξ /R) I (integral from t to 0) e t/RC Remember that I (integral) eax dx = (1/a) eax o Therefore, q = ( ξ /R)[RC][e t/RC ] (evaluated from t to 0) o q = ξ C [ e t/RC1] Or similarly. .. q(t) = C ξ[ 1 e t/RC ] q(t) = C ξ[ 1 e t/ τ ], where τ is the time constant which equals RC. • See Figure 2814 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 et/RC • d/dt [q(t) = q max et/RC ] oi = dq/dt = q max (1/RC)( et/RC ) o So, i = ( q max /RC)( et/RC ) o i = i o et/RC • See problem 72 from Chapter 28 of the text...
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 Spring '07
 MAHLON,GREGORYDA
 Magnetism, RC Circuits

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