in-class problems
DESCRIPTION
In-Class Problems. Sketch the following functions: a) x(t) = 3sin(40 p t) for 0≤ t ≤ 0.2 sec b) z(t) = 10e -4t for 0≤ t ≤0.5 sec What is ? What is ? What is ?. Capacitor-Resistor Circuits. In-class activity: - PowerPoint PPT PresentationTRANSCRIPT
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In-Class Problems
1. Sketch the following functions:a) x(t) = 3sin(40t) for 0≤ t ≤ 0.2 secb) z(t) = 10e-4t for 0≤ t ≤0.5 sec
2. What is ?
3. What is ?
4. What is ?
)40sin(3)( tdt
dtf
tedt
dty 410)(
tdt
dtg 6)(
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Capacitor-Resistor Circuits
• 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).
V112 V
R1
1k
C11uF
J1
Key = Space
1
2
4
0
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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
V112 V
R1
1k
C11uF
J1
Key = Space
1
2
4
0
Vs
-vc
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Class Activity
• What are the units of RC in vc(t) = Vse-t/RC ?
• Since vc(t) = Vse-t/RC and , what is i(t) in terms of Vs, R and C?
• What do vc(t) and i(t) look like on a graph?
cdvi t Cdt
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Step Response: RC Time Constants
• Now, what happens when the switch moves the other way?
V112 V
R1
1k
C11uF
J1
Key = Space
1
2
4
0
The response of the capacitor voltage will be to charge up to the supply voltage.
Vs
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Vc Response to Constant Voltage Vs
• The voltage across the capacitor will rise and asymptotically approach Vs
How can we describe this mathematically?
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Kirchhoff’s voltage loop law S C RV v v
Ohm’s law across resistor Rv iRSubstituting for VR gives s cV v iR
Analysis of RC Circuits
V112 V
R1
1k
C11uF
J1
Key = Space
1
2
4
0
Vs
vc
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Substitute in for i
From previous pageS CV v iR
CS C
dvV v RC
dt
Analysis of RC Circuits
CdvCdt
V112 V
R1
1k
C11uF
J1
Key = Space
1
2
4
0
Vs
vc
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Analysis of RC Circuits
• The equation is called a differential equation.
CS C
dvV v RC
dt
• The solution is of the form:
where is defined as the time constant
1t
C Sv V e
RC
= the circuit time constant, in seconds if and only if C = the total (connected) capacitance Farads R = the total (connected) resistance Ohms
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Team Activity
CS C
dvV v
dt
• Substitute
into the equation
to show that LHS = RHS
1t
C Sv V e
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Show that when t is 5 times the time constant, ,the capacitor voltage is 99.33% of the peak voltage.
1t
C
S
ve
V
Team Activity
100SV V
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Kirchhoff’s voltage loop law ?
Ohm’s law across resistor?
Substituting for VR gives?
Team Activity – Discharge Process
V112 V
R1
1k
C11uF
J1
Key = Space
1
2
4
0
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CC
dvv
dt• From previous activity the equation
t
C Sv V e
Team Activity
and show that LHS = RHS
• Substitute
into the above equation
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Rectangular Wave
• If you repeatedly switch between the battery and the short you are effectively applying a rectangular time pulse to the RC circuit.
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Rectangular Wave Response
• The voltage across the capacitor will behave as below in response to such a wave: