Transcript Document

Admin:
• Assignment 7 is complete
• Assignment 8 is posted
• Due Monday (27th)
• Phasor operations and AC analysis
• Second mid-term date is set:
• Wednesday May 6th
https://www.youtube.com/watch?v=BhMSzC1crr0
Amplitude A=16.67 Volts
v(t)
is(t)
ω=2 rads/sec = 2πf
f=1/π Hz = 0.318 Hz
T=π secs = 3.1416 secs
Phase offset φ is
56.3° ≈ 1 radian
=T/2π= 0.5 secs
If ω were different, the impedances would change –
producing a different amplitude and phase shift
Vpeak=16.67 V
Ipeak =10 Amps
VRMS=11.8 V
IRMS=7.1 Amps
Amplitude A=16.67 Volts
T=3.14s
=2π rads
Δt
φ
v(t)
is(t)
ω=2 rads/sec = 2πf
f=1/π Hz = 0.318 Hz
T=π secs = 3.1416 secs
Phase offset φ is
56.3° ≈ 1 radian
Δt=1×T/2π= 0.5 secs
If ω were different, the impedances would change –
producing a different amplitude and phase shift
AC circuit analysis: some things to note
• You need to be familiar with complex number operations
• You will need to be able to solve simple systems of equations with
complex numbers (2 equations, 2 unknowns)
• ω is the angular frequency, in radians/sec
– It is not the same as the frequency, f, which is in Hz
–
ω = 2πf
• Take care not to mix/confuse radians and degrees
• Don’t forget the “multiply by j/j” trick.
• Numbers with no real (or imaginary) part can still be written as
complex numbers.
• Come to office hours if any of this makes no sense. It’s not difficult,
but it can be confusing - I can help!
• I will work through some more examples in the next few lectures –
but just one per class, while I continue with other material, to prevent
our brains melting.
I’ll go through this one next class
Or try it yourself first…
Transient Circuits
t<0
What happens in the
transition region between
two steady states of a
circuit like this?
What happens when the
switch is closed?
t=∞
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Transient Circuits
t<0
Before the switch is
closed (at t<0) – no
current is flowing, so
there is no charge on C,
and no voltage across it.
What happens when the
switch is closed?
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Transient Circuits
What happens when the
switch is closed (t=0)?
qC = CvC
iC = C
dvC
dt
A sudden change in vC
would correspond to an
infinite current – which is
not physical.
There is no sudden jump of
voltage on the capacitor
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Transient Circuits
So what really happens?
Apply KVL:
e - vR - vC = 0
e - iR - vC = 0
As the switch is closed vC=0,
so
e - iR = 0
e
i=
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R
So at t=0, the capacitor acts just like a
piece of wire (a short). The current is
maximum, the capacitor voltage is 0.
Transient Circuits
We can see this from an
impedance perspective as
well:
XC =
1
wC
As the switch is closed
ω=∞, so XC=0
So at t=0, the capacitor acts
just like a piece of wire
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What happens at t=∞?
At t=∞, the Capacitor is
fully charged.
qC = constant
iC =
dqC
=0
dt
Apply KVL:
So at t=∞, the current
is zero, the voltage
maximum.
At t=∞
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e - iR - vC = 0
vC = e
qC = CvC = Ce
What happens in between t=0 and
t=∞?
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Charging a capacitor
When the switch is
closed, the
capacitor will begin
to charge. As it
does, the voltage
across it increases,
and the current
through the resistor
decreases.
But not linearly…
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Charging a capacitor
During Charging:
e - vR (t) - vC (t) = 0
dQC
dv
iC =
=C C
dt
dt
v (t) e - vc (t)
iC = iR = R =
R
R
dv
e - vc (t)
so: C C =
dt
R
dvC
dt
=
e - vc (t) RC
Integrate both sides using:
Gives:
1
1
dx

 ax  b a ln(ax  b)
t
ln(  vC (t ))  
A
RC
  vC (t )  e e
A

t
RC

 Be
Boundary condition: at t=0, vC(t)=0 so:
  vc (t )  e

t
RC
vC (t )   (1  e

t
RC
)
t
RC
  Be0  B
Charging a capacitor
Solution is only true for that particular circuit (Voltage source plus
resistor), but more complicated circuits can be reduced to this using
Thevenin's Theorem
0.37
Time constant τ=RC. Time needed to charge capacitor to 63% of full charge
Larger RC means the capacitor takes longer to charge
Larger R implies smaller current flow
The larger C is, the more charge the capacitor can hold.
Analyse the units…
Summary
To find the voltage as a function of time, we
write the equation for the voltage changes
around the loop:
Since Q = dI/dt, we can integrate to find the
charge as a function of time:
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The voltage across the capacitor is VC = Q/C:
The quantity RC that appears in the exponent
is called the time constant of the circuit:
Analyse the units…
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The current at any time t can be found by
differentiating the charge:
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