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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) e-ax dx = -(1/a) e-ax o Therefore, q = ( ξ /R)[-RC][e- t/RC ] (evaluated from t to 0) o q = -ξ C [ e- t/RC-1] 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 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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- Spring '07
- MAHLON,GREGORYDA
- Magnetism, RC Circuits, Trigraph, Electric charge, RC circuit, Kirchhoff's circuit laws, Exponential decay, Gustav Kirchhoff