Kirchoff's Laws and Capacitors. Vo ! q. C ! iR = 0. " i t( ) +. q t( ). RC. = Vo. R. " dq
dt. +. 1. RC. #. $%. &. '( q = io. " q t( ) = qmax 1! e!t / RC. (. ) and. " i t( ) = dq dt. =.
Charging characteristics of initially uncharged capacitor.
q ! iR = 0 C q ( t ) Vo " i( t) + = RC R dq # 1 & " +% q = io $ RC (' dt
One time constant is defined as ! = RC . Putting that value into our charging relationship yields:
R
Kirchoff’s Laws and Capacitors
( (
q ( t=! ) = q max 1 " e " t /RC
C
Vo
Vo !
(
" q ( t ) = q max 1 ! e ! t /RC
)
= q max 1 " e " RC/RC
)
q(t)
1& # = q max % 1 " 1 ( $ e ' throw at t=0
)
= .63q max
q max
Evidently, the charge on the capacitor is equal to 63% of its total possible charge after charging for a period equal to one time constant, or a time equal to “RC.”
and
dq q = max e ! t /RC dt RC " i ( t ) = io e ! t /RC " i( t) =
There is a slight subtlety in the relationship between the charge flow (i=dq/dt) and the amount of charge (q) on the capacitor’s plate at a given instance. I’m not making a big deal about this as the bottom line is good and you have enough to worry about without the added burden of an overly intricate derivation.
.63q max
t=!
This is depicted on the graph to the right.
t
( =RC)
1.
3.
The charging relationship is graphed below:
After two time constants:
( (1 " e
q ( t=! ) = q max 1 " e " t /RC q(t)
= q max
)
"2RC/RC
)
1& # = q max % 1 " 2 ( $ e '
q max
q(t)
= .87q max
(
q ( t ) = q max 1 ! e
! t /RC
)
q max Evidently, the charge on a capacitor will be equal to 87% of its total possible charge after charging for a period equal to two time constants.
t
This is depicted on the graph to the right.
.87q max
t = 2!
t
( =2RC)
2.
4.
It should be noted that a capacitor’s DISCHARGING relationship is a mirror image of the charging relationship. That is, after one time constant, the discharging capacitor will have dumped 63% of its charge, After two time constants worth of time, it will have dumped 87% of its charge. Those relationships are shown in the discharging graph shown to the right.
q(t)
io
.37q max
q ( t ) = q max e ! t /RC
dumped 63% of q max , 37% left dumped 87% of q max , 13% left
t=!
( =RC)
t = 2!
t
( =2RC) 5.
Something similar happens with the i(t) current versus time graph for both a charging and discharging capacitor circuit. The analysis there yields: io
i ( t ) = io e ! t /RC = io e ! RC/RC " 1% = io $ 1 ' #e &
i ( t ) = io e ! t /RC
.37io
= .37io Evidently, the current in a charging and discharging circuit will diminish to 37% of its maximum (the maximum occurring right at the start of the charging or discharging process) over a time equal to “RC”--one time constant.
