RC Circuits-Charging a Capacitor

RC Circuits-Charging a Capacitor - • Now i = dq/dt o...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 202 Tuesday, March 16, 1999 Announcements: Exam Thursday! Material Covered in today's lecture WILL NOT be on midterm 2 Lecture notes: RC Circuits Charging a Capacitor At time t = 0, the switch is closed. o At some later time, we can use Kirchhoff's rule, ξ - iR - (q/C) = 0 ( equation A) Note that at t = 0, q = 0. Therefore, ξ - i o R = 0 . o So, i o = ξ/ R . This is the current at t = 0. o 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. o In Equation A, for very large t, ξ - (q max /C) = 0. Therefore, q max = C ξ. To determine the time dependence of i and q, (d/dt) [ ξ - iR - (q/C) ] o = 0 - R(di/dt) - (1/C)(dq/dt) o But, i = dq/dt o Therefore, R(di/dt) + (1/C)i =0 o 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
ln (i/i o ) = - t/ RC i(t) = i o e - t/RC where e is the exponential not an electron o Therefore, i (t) = ( ξ /R)e - t/RC Current drops off exponentially
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Page1 / 3

RC Circuits-Charging a Capacitor - • Now i = dq/dt o...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online