Transcript Demonstration - Faculty Pages

In-Class Problems
1. Sketch the following functions:
a) x(t) = 3sin(40pt) for 0≤ t ≤ 0.2 sec
b) z(t) = 10e-4t for 0≤ t ≤0.5 sec


d
 4t
y
(
t
)

10
e
2. What is
?
dt
3. What
d
is f (t )  dt 3 sin( 40pt ) 
d
4. What is g (t )  6t  ?
dt
?
Capacitor-Resistor Circuits
J1
2
1
Key = Space
V1
12 V
R1
1k
4
C1
1uF
0
• In-class activity:
• In pairs build this circuit in Multisim.
• Look at the voltage across the capacitor on the
oscilloscope.
• Describe what you see when the switch moves
between positions (let the switch stay in each
position until the capacitor voltage stops changing).
Discharge of a Capacitor
Through a Resistor
• In the following circuit, when the switch
moves from the battery to the wire
connected to ground, the voltage across
the capacitor is vc(t) = Vse-t/RC
J1
2
1
Key = Space
Vs
R1
1k
4
V1
12 V
vc
0
C1
1uF
-
Class Activity
• What are the units of RC in vc(t) = Vse-t/RC ?
dv
c
-t/RC
i
t

C
• Since vc(t) = Vse
and  
, what
dt
is i(t) in terms of Vs, R and C?
• What do vc(t) and i(t) look like on a graph?
Step Response: RC Time Constants
• Now, what happens when the switch moves the
other way?
J1
2
1
Key = Space
R1
1k
4
V
V1
12 Vs
C1
1uF
0
The response of the capacitor voltage will be to
charge up to the supply voltage.
Vc Response to Constant Voltage Vs
• The voltage across
the capacitor will rise
and asymptotically
approach Vs
How can we describe this mathematically?
Analysis of RC Circuits
J1
2
1
Key = Space
Vs
R1
1k
4
V1
12 V
vc
C1
1uF
0
Kirchhoff’s voltage loop law
VS  vC  vR
Ohm’s law across resistor
vR  iR
Substituting for VR gives
Vs  vc  iR
Analysis of RC Circuits
J1
2
1
Key = Space
Vs
R1
1k
4
V1
12 V
vc
C1
1uF
0
From previous page
VS  vC  iR
dvC
Substitute C
in for i
dt
dvC
VS  vC  RC
dt
Analysis of RC Circuits
dvC
• The equation VS  vC  RC
is called a differential
dt
equation.
t
 


• The solution is of the form: vC  VS 1  e 


where   RC is defined as the time constant
 = the circuit time constant, in seconds if and only if
C = the total (connected) capacitance Farads
R = the total (connected) resistance Ohms
Team Activity
t
 

• Substitute vC  VS 1  e  


dvC
into the equation VS  vC  
dt
to show that LHS = RHS
Team Activity
VS  100V
vC
 1 e
VS

t

Show that when t is 5 times the time constant,  ,
the capacitor voltage is 99.33% of the peak voltage.
Team Activity – Discharge Process
J1
2
R1
1
Key = Space
1k
V1
12 V
4
C1
1uF
0
Kirchhoff’s voltage loop law ?
Ohm’s law across resistor?
Substituting for VR gives?
Team Activity
• From previous activity the equation
• Substitute
vC  VS e

t

into the above equation
and show that LHS =
RHS
dvC
vC  
dt
Rectangular Wave
• If you repeatedly switch between the battery and
the short you are effectively applying a
rectangular time pulse to the RC circuit.
Rectangular Wave Response
• The voltage across the capacitor will behave as
below in response to such a wave: