REVIEW-Transients

REVIEW-Transients - ENGR 210 REVIEW Transients Page 1...

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THINGS YOU SHOULD KNOW ABOUT TRANSIENTS (Lectures 20-28) 1. i-v relationships for capacitors and inductors 2. Energy in capacitors and inductors. 3. Time constants. 4. Computing initial and final values. 5. Initial/final value theorem. 6. Equivalent circuits as seen by inductor/capacitor. PAGE DISCHARGING MULTIPLE CHARGED CAPACITORS 2-4 BASIC R-C CIRCUIT 5-7 THEVENIZING R-C CIRCUITS 7-11 SWITCHED CIRCUITS 12-17 TRANSIENT CIRCUITS W/DEPENDENT SOUCES 18-19 MUTUAL INDUCTANCE R-L CIRCUITS 20-22 MULTIPLE SWITCHED CIRCUITS 23-25 PULSES IN R-C AND R-L CIRCUITS 26-29 OP-AMP INTEGRATORS AND DIFFERENTIATORS 30-31 ENGR 210 REVIEW Transients Page 1
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ENGR 210 HW9 Solutions Page 3 Problem 2 Switched Capacitors At the time the switch is closed in the circuit shown below, the capacitors are charged as shown. 250k ± v o + - v 1 + - v 2 + - 30V + - 5V + - 6 μ F 1 μ F 2 μ F (a) Find v o t () for t ± 0 + . SOLUTION: This problem is very similar to Example 7.4 which was worked in class. You need to start by finding the equivalent capacitances to determine the time constant. Combining the 2 ² f and the 1 ² f capacitor gives a 3 ² f capacitor with an initial voltage of 30 volts. Combining the 3 ² f and the 6 ² f capacitor gives 1 C eq = 1 6 ± 10 ² 6 + 1 3 ± 10 ² 6 or C eq =2 ² f . The initial voltage v o 0 is gotten by combining v o 0 = v 1 0 + v 2 0 = ± 5 + 30 = 25 volts. Since there are no sources of energy the final voltage across the equivalent capacitor will be zero. The time constant is given as ± = RC eq = 250 ± 10 3 2 ± 10 ² 6 = 0.5 seconds. Using the initial-final value theorem for the equivalent capacitor v C t = v C ± + v C 0 ² v C ± ³ ´ µ · ¸ e ² t ± t ¹ 0 v C t = 0 + 25 ± 0 ² ³ ´ µ · e ± t 0.2 = 25 e ± 5 t which is also v o t (b) What percentage of the total energy initially stored in the three capacitors is dissipated in the 250k ³ resistor? SOLUTION: ENGR 210 REVIEW Transients Page 2
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ENGR 210 HW9 Solutions Page 4 The initial energy is simply the sum of the energies in the individual capacitors. w e 0 () = 1 2 C 1 v 1 2 0 + 1 2 C 2 v 2 2 0 + 1 2 C 3 v 2 2 0 w e 0 = 1 2 6 ± 10 ² 6 ² 5 2 + 1 2 2 ± 10 ² 6 30 2 + 1 2 1 ± 10 ² 6 30 2 w e 0 = 75 ± 10 ² 6 + 900 ± 10 ² 6 + 450 ± 10 ² 6 = 1425 ± J You can now calculate p diss t = v o 2 t 250 k and integrate it to get the dissipated power. However, there is a simpler method. See p. 276 at the end of example 7.4. The energy stored in the equivalent capacitor is the energy delivered to the resistor, i.e., the power dissipated. For this equivalent capacitor we then have w diss = 1 2 C eq v o 2 t = 1 2 2 ± 10 ² 6 25 2 = 625 ± 10 ² 6 = 625 ± J The percentage energy dissipated is then % dissipated = w diss w e 0 = 625 1425 = 43.8% (c) Find v 1 t for t ± 0. SOLUTION: You need to work from the capacitor current to find the voltage across the individual capacitors. The current through the resistor is i o t = v o t 250 ± 10 3 . This is the same as the capacitor current but in the opposite direction. The current through the 6 ² f capacitor is then i C 1 t = 25 e ² 2 t 250 ± 10 3 = 100 ± 10 ² 6 e ² 2 t . The voltage across this capacitor is then given by integrating i C 1 t = C 1 dv 1 dt or v 1 t = v 1 0 + 1 C 1 i C 1 x 0 t ± dx = ² 5 + 1
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This document was uploaded on 04/18/2011.

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REVIEW-Transients - ENGR 210 REVIEW Transients Page 1...

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