General RC Solution

Every current or voltage (except the source voltage) in
an RC circuit has the following form:

(
t

t
)
/(
RC
)



0
x( t )  x f   x( t0 )  x f e


x represents any current or voltage
 t0 is the time when the source voltage switches
 xf is the final (asymptotic) value of the current or voltage
All we need to do is find these values and plug in to solve
for any current or voltage in an RC circuit.

Solving the RC Circuit
We need the following three ingredients to fill in our
equation for any current or voltage:
x(t0+) This is the current or voltage of interest just after
the voltage source switches. It is the starting point of our
transition, the initial value.
 xf
This is the value that the current or voltage
approaches as t goes to infinity. It is called the final value.
 RC This is the time constant. It determines how fast the
current or voltage transitions between initial and final value.

Example
R1 = 10 kW
t=0
I
+
Vs = 5 V 
Find the current I(t).
R2 = 5 kW
C = 1 mF
Finding the Initial Condition
To find x(t0+), the current or voltage just after the switch, we use
the following essential fact:
Capacitor voltage is continuous; it cannot jump when a
switch occurs.
So we can find the capacitor voltage VC(t0+) by finding VC(t0-),
the voltage before switching.
We can assume the capacitor was in steady-state before
switching. The capacitor acts like an open circuit in this case, and
it’s not too hard to find the voltage over this open circuit.
We can then find x(t0+) using VC(t0+) using KVL or the capacitor I-V
relationship. These laws hold for every instant in time.
Example: Initial Condition
R1 = 10 kW
t=0
I
+
Vs = 5 V 





R2 = 5 kW
C = 1 mF
We first find capacitor voltage right after the switch, (at t=0+)
and use it to find the current I at t=0+.
To do this, we look at the circuit before switching, because the
capacitor voltage will remain the same after switching.
Assuming the circuit has been “unswitched” for a long time, the
capacitor acts like an open circuit connected to a resistor.
The capacitor voltage before switching (at t=0-) is 0 V.
The capacitor voltage after switching (at t=0+) is 0 V.
Example: Initial Condition
R1 = 10 kW
+ 5V +
Vs = 5 V 
R2 = 5 kW
t=0
I
+
0V
_
C = 1 mF

We know the capacitor voltage is 0 V right after the switch.

By KVL, the resistor gets all 5 V that the source puts out.

So by Ohm’s law, I(0+) is 5 V / 10 kW = 0.5 mA
Finding the Final Value
To find xf , the asymptotic final value, we assume that the circuit will
be in steady-state as t goes to infinity.
So we assume that the capacitor is acting like an open circuit. We
then find the value of current or voltage we are looking for using this
open-circuit assumption.
Here, we use the circuit after switching along with the open-circuit
assumption.
When we found the initial value, we applied the open-circuit
assumption to the circuit before switching, and found the capacitor
voltage which would be preserved through the switch.
Example: Final Value
R1 = 10 kW
t=0
I
+
Vs = 5 V 



R2 = 5 kW
C = 1 mF
After the circuit has been switched for a long time, the capacitor
will act like an open circuit.
Then no current can flow—eventually, I goes to zero.
If = 0 A
Finding the Time Constant
It seems easy to find the time constant: it equals RC.
But what if there is more than one resistor or capacitor?
R is the Thevenin equivalent resistance with respect to the
capacitor terminals.
Remove the capacitor and find RTH. It might help to turn off the
voltage source. Use the circuit after switching.
We will discuss how to combine capacitors in series and in parallel
in the next lecture.
Example: Time Constant
R1 = 10 kW
t=0
I
+
Vs = 5 V 




R2 = 5 kW
C = 1 mF
How long does it take for the current to converge after
switching?
Look at the circuit after switching to find the time constant.
The 5 kW resistor is not “in” the circuit after the switch.
R = 10 kW, C = 1 mF
so t = 10 s
Example: Final Answer
R1 = 10 kW
t=0
I
+
Vs = 5 V 

R2 = 5 kW
Plugging in, we get:
I(t) = If + (I(0+) – If) e-t/RC
I(t) = 0 + (0.5 mA – 0) e-t/10 = 0.5 e-t/10 mA
C = 1 mF
Alternative Method
Instead of finding these three ingredients for the generic
current or voltage x, we can
 Find the initial and final capacitor voltage (it’s easier)
 Find the time constant (it’s the same for everything)
 Form the capacitor voltage equation VC(t)
 Use KVL or the I-V relationship to find x(t)
Example: Alternative Method
R1 = 10 kW
t=0
I
+
Vs = 5 V 
R2 = 5 kW

The initial capacitor voltage was VC(0+) = 0 V.

The time constant is t = 10 s.
C = 1 mF

The final value of the capacitor voltage, when the capacitor has
gone to an open circuit, is VC,F = 5 V.
Plug in to get V (t) = 5 – 5 e-t/10 V.

Since I = C dV/dt,

C
I(t) = (10-3)((-5)/(-10) e-t/10 A
Series and Parallel Capacitors

Capacitors combine the opposite way as compared to resistors.
C1
C1
C2
C2
C3
C3
=
=
C1 + C2 + C3
(1/C1 + 1/C2 + 1/C3)-1
Hard Example
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
2 F
1 kW


Find V1(t) and V2(t).
Find the energy absorbed by the resistors for t > 0.
Initial Conditions
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
2 F
1 kW

Consider the circuit before the switch—it had been that way for
a long time. Assume second capacitor is discharged at t=0.
Time Constant
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
1 kW


How long does it take to converge after switch?
What are R and C?
2 F
Final Value
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
2 F
1 kW



After a while, both capacitors are open circuits.
They can have nonzero voltage, but voltages must be equal.
Equate V = Q/C for capacitors to gain insight into final value…
Solutions

Energy is time integral of power. Find P = VI = R(I2).
DRAM





Dynamic Random Access Memory (DRAM) cells are really just
capacitors. Each capacitor is one bit.
The circuit we just saw simulates the writing and reading of a
DRAM cell (capacitor 1).
A logic gate that needed to use the contents of the DRAM cell
would be represented by an RC circuit, and the DRAM capacitor
would need to charge the logic gate’s natural capacitance.
DRAM capacitors are connected to electronic switches, not
physical switches. So there is always a little current flowing.
This drains the charge from the DRAM capacitors. They need
to be “refreshed” periodically. Hence the term dynamic.