Laplace Circuit Analysis Solution

Laplace Circuit Analysis Solution - EE 2011 Practice...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 2011 Practice Problems Laplace Circuit Analysis Q1 This circuit has been in the steady state for a long time before the switch closes at t = 0. 1. Draw the s-domain circuit for t > 0. 2. Calculate I L (s). 3. Calculate i L (t) for t > 0. I L (s) = 24/s – 96,000/(s(s + 1000)(s + 4000)) = 24(s + 5000)/((s + 1000)(s + 4000)) = A/(s + 1000) + B/(s + 4000) A = 32, B = -8 i L (t) = 32 exp(-1000 t) – 8 exp(-4000 t) 16V TCLOSE = 0 24A 125uF 2mH 3.333 i L Q2 This circuit has been in the steady state for a long time before the switch closes at t = 0. a) Draw the s-domain circuit for t > 0. a) Find I L (s) b) i L (t) for t > 0. 1. Draw the circuit in the Laplace domain, with the switch closed, and with the DC currents and voltages tagged with 1/s. Remember to include sources for the initial inductor current and capacitor voltage. 2. Replace the 24 A source and the two 2 ohm resistors with their Thevenin equivalent (48/s volts and 4 ohms) 3. Write a single KCL equation for VL to solve the circuit. (VL – 48/s)/4 + 24/s + VL s/8000 + VL 500/s = 0 Solve for VL VL = -96,000/(s^2 + 2000 s + 4*10^6) IL = 24/s + VL 500/s = (24 s^2 + 48000 s + 144*10^6)/(s(s^2 + 2000 s + 4*10^6)) = A/s + (Bs + C)/(s^2 + 2000 s + 4*10^6) A = 36 B = -12 C = -24000 IL = 36/s + (-12s - 24000)/(s^2 + 2000 s + 4*10^6) IL = 36/s -12(s + 1000)/((s + 1000)^2 + 1732^2) – 6.93(1732)/((s + 1000)^2 + 1732^2) iL(t) = (36 – 12 exp(-1000 t) cos(1732 t) – 6.93 exp(-1000 t) sin(1732 t)) u(t) 2 Ω 2 Ω 125 ufd 2 mH 16 V 24 A t = 0 i L v C Q3 The circuit below had been running for a long lime when at time t = 0 the leftmost capacitor, C 1 , failed. Surprisingly, the capacitor failed short, that is, it became a short circuit. All the energy in C 1 was dissipated in the destruction of the capacitor and we are not interested in it. Using Laplace techniques, calculate v O (t) and i g (t) for positive time. Draw two circuits, the first with vg and L1 and the second with L2, C2, and Rload (along with initial inductor currents and capacitor voltages)....
View Full Document

This note was uploaded on 04/09/2008 for the course EE 2011 taught by Professor Imbertson during the Spring '08 term at Minnesota.

Page1 / 16

Laplace Circuit Analysis Solution - EE 2011 Practice...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online