Download Document

Admin:
• Assignment 7 is posted
• Due Monday (20th)
• Hints:
• For capacitors in series, the charge is the same
• For capacitors in parallel the voltage is the same
• Second mid-term date is set:
• Wednesday May 6th
http://apod.nasa.gov/apod/ap110102.html
https://www.youtube.com/watch?feature=player
_embedded&v=nTZQn19EwAU
Impedance is the complex form of a component’s resistance
Capacitative Impedance
Inductive Impedance
Resistive Impedance
1
ZC =
jwC
Z L = jw L
ZR = R
What if there is more than one component?
AC circuit analysis
• Procedure to solve a problem
– Identify the source sinusoid and note the frequency
– Convert the source(s) to complex/phasor form (you can ignore the ωt component
at this point)
– Represent each circuit element by it's AC impedance. Impedances add like
resistors.
– Solve the resulting phasor circuit using standard circuit solving tools (equivalent
impedances, voltage/ current divider, KVL,KCL,Mesh, Thevenin etc.)
Impedances combine like restances in a DC circuit.
– Find the real part of the solution, and write it as a function of time.
This circuit contains an AC current source is(t)=10cos2t Amps.
Use the generalized Ohm’s law to find the voltage, v(t).
Note the
frequency
Note: the regular number “1” is the complex number “1+0j ”
j2 = -1
Now switch to
phasor form to
make life easy
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