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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
