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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=∞ Copyright © 2009 Pearson Education, Inc. 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? Copyright © 2009 Pearson Education, Inc. 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 Copyright © 2009 Pearson Education, Inc. 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= Copyright © 2009 Pearson Education, Inc. 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 Copyright © 2009 Pearson Education, Inc. 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=∞ Copyright © 2009 Pearson Education, Inc. e - iR - vC = 0 vC = e qC = CvC = Ce What happens in between t=0 and t=∞? Copyright © 2009 Pearson Education, Inc. 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… Copyright © 2009 Pearson Education, Inc. 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 Be 0 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: Copyright © 2009 Pearson Education, Inc. 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… Copyright © 2009 Pearson Education, Inc. The current at any time t can be found by differentiating the charge: Copyright © 2009 Pearson Education, Inc.