ECE 231 -9

ECE 231 -9 - ECE-231 Circuits and Systems I Spring 2011...

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Unformatted text preview: ECE-231 Circuits and Systems I Spring 2011 Session 9 Quiz 3: April 12 (3 weeks from today) Professor Stewart Personick Office: ECEC Room 321 [email protected] Initial current at t=0s is 0A Target current for 0s<t<2s is1000A Target current for 2s<t is 0A 1000A Initial current at t=2s is 865A HW#8 Problem 1 HW#8 Problem 2:Mutual Inductance +- v(t) = 20 sin(2000 π t) Volts a b c d Assume γ =1, and therefore: vab(t) = N2/N1 x vcd(t) i2(t) = N1/N2 x i1(t) N1 = 1000 turns Magnetic field lines i1(t ) = C sin(2000 π t) Amperes i2(t) = B sin(2000 π t) Amperes = Vab(t) / 8 Ω R= 8 Ω Calculate: A, B, and C What is v(t)/i1(t) (in Ohms)? vab(t) = A sin(2000 π t) Volts N2 = 50 turns A = 20 x 50/1000 = 1 B = A /8 = 1/8 = 0.125 C = 0.125 x 50/1000= 0.00625 v(t)/i1(t) = 20V/.00625A = 3200 Ohms Substituting Complex Exponentials for Sine and Cosine Waveforms If i(t) = A e j2 π ft and i(t) flows through a resistor, then the voltage produced across the resistor is: VR(t) = R x i(t) = R x A e j2 π ft Substituting Complex Exponentials for Sine and Cosine Waveforms If i(t) = A e j2 π ft and i(t) flows through an inductor, then the voltage produced across the inductor is: VL(t) = L x d[ i(t) ]/dt = L x j 2 π f x A e j2 π ft Substituting Complex Exponentials for Sine and Cosine Waveforms If i(t) = A e j2 π ft and i(t) flows through a capacitor, then the voltage produced across the capacitor is {since i(t) = C d[VC(t)]/dt}: VC(t) = (1/C) x the integral of i(t) dt = (1/C) x (1/ j 2 π f) x A e j2 π ft Using complex exponentials to solve for currents and voltages in circuits Circuit containing resistors, inductors and capacitors +- Vs(t) = A e j2 π ft = A cos (2 π ft) + j A sin (2 π ft) Using complex exponentials to solve for currents and voltages in circuits Circuit containing resistors, inductors and capacitors +- Vs(t) = A e j2 π ft = A cos (2 π ft) + j A sin (2 π ft) First solve for the node voltages and mesh currents using Vs(t) : For example: Vab(t) = Γ e j2 π ft and i1(t) = Η e j2 π ft Then take the real part of each result to obtain: Vab(t) = Re { Γ e j2 π ft} and i1(t) =Re { Η e j2 π ft} Vs(t) Using sine waves to solve for currents and voltages in RLC networks (example) +- Vs(t) = A e j2 π ft a Vs(t) = Vab(t ) + Vbc(t) + Vcd(t) [i.e. Kirchhoff’s voltage law] A e j2 π ft = j2 π f x L x Λ x e j2 π ft + 1/[j2 π f x C] x Λ x e j2 π ft + R x Λ x e j2 π ft A = j2 π f x L x Λ + 1/[j2 π f x C] x Λ + R x Λ b i(t) = Λ x e j2 π ft- + Vab(t) c d Solve for the unknown: Λ R C L Using sine waves to solve for currents and voltages in RLC networks (example) +- Vs(t) = A e j2 π ft a Vs(t) = Vab(t) + Vbc(t) + Vcd(t) A e j2 π ft = j2 π f x L x Λ x e j2 π ft + 1/[j2 π f x C] x Λ x e j2 π ft + R x Λ x e j2 π ft A = j2 π f x L x Λ + 1/[j2 π f x C] x Λ + R x Λ b i(t) = Λ x e j2 π ft- + Vab(t) c d R C L Using sine waves to solve for currents and voltages in RLC networks (example) +- a Vs(t) = Vab(t) + Vbc(t) + Vcd(t)...
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This note was uploaded on 03/26/2012 for the course ECE 231 taught by Professor Pietrucha during the Spring '08 term at NJIT.

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ECE 231 -9 - ECE-231 Circuits and Systems I Spring 2011...

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