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– iC = 0 +
vC = VS
– after a sufﬁciently long time
5. Eventually, the capacitor voltage will increase until it is equal to
VS. At that point, the current drops to zero. From that time nothing
changes again. Now, it doesn’t matter if the switch is closed or open.
This is a transient effect. Initially, the capacitor voltage was at some ﬁxed,
steady value and the capacitor current was zero. When the switch closed,
the current jumped up to a value determined by the difference between
the source voltage, the initial capacitor voltage and the resistance
connecting them. The current charged the capacitor to increasingly
higher voltage. When the capacitor voltage reached the source voltage,
the current dropped to zero, and the circuit to a static (quiescent) state
with the capacitor at a new, higher voltage with no current ﬂowing.
EE 442 RC and RL transients – 5 R
VS +
– iR iC(t) C+
vC(t)
– Now, let’s do the math. For t > 0 (after the switch has closed):
KCL iC = iR
V S − v C ( t)
iR =
R
dvC (t)
iC = C
dt
dvC (t)
V S − v C ( t)
C
=
dt
R This is a differential equation. Fortunately, it is probably the ﬁrst one you
learned to do in your ﬁrst diff. eq. class.
EE 442 RC and RL transients – 6 dvC
VS − v C
C
=
dt
R
dvC
dt
=
VS − v C
RC Solve it by integrating both sides.
vC (t)
VCi
dvC VS −
vC = t
o dt
RC The right side is time, and we integrate from t = 0 to t. (As usual, use a
dummy variable in the integrand.)
The left side is voltage, and the integration limits must match those on the
time side. Therefore, we integrate from vC(t = 0) = VCi to vC(t). EE 442 RC and RL transients – 7 Carrying out the integration:
vC (t )
t
t
− ln (VS − vC )
=
RC o
VCi and manipulating the result t
− ln [VS − vC (t)] + ln [VS − VCi ] =
RC
VS − VCi
t
− ln
=
V S − v C ( t)
RC
VS − VCi
= exp
V S − v C ( t) t
RC t
VS − vC (t) = [VS − VCi ] exp −
RC
t
vC (t) = VS − [VS − VCi ] exp −
RC
EE 442
RC and RL transients – 8 t
vC (t) = VS − [VS − VCi ] exp −
RC The quantitative result matches the earlier qualitative argument exactly. At...
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 Fall '09
 Volt

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