This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: www.electrical.net.tc 2010 www.electrical.net.tc Page 1 PROBLEMS Q#5.1: In the network of the figure, the switch K is closed at t = 0 with the capacitor uncharged. Find values for i, di/dt and d 2 i/dt 2 at t = 0+, for element values as follows: V 100 V R 1000 C 1 F + R V C  Switch is closed at t = 0 (reference time) We know Voltage across capacitor before switching = V C (0) = 0 V According to the statement under Q#5.1. V C (0+) = V C (0) = 0 V V 100 i C (0+) = i(0+) = = = 0.1 Amp. R 1000 Element and initial condition Equivalent circuit at t = 0+ C Sc Switch Drop Rise i(0+) Short circuit Drop Applying KVL for t 0 Sum of voltage rise = sum of voltage drop www.electrical.net.tc 2010 www.electrical.net.tc Page 2 1 V = iR + idt C Differentiating with respect to „t‟ di i R + = 0 dt C di(0+) i(0+) = [eq. 1] dt CR By putting the values of i(0+), C & R di(0+) (0.1) = dt (1 F)(1 k ) di(0+) = 100 Amp/sec dt Differentiating eq. 1 with respect to „t‟ d 2 i(0+) di(0+) 1 = dt 2 dt CR Putting the corresponding values d 2 i(0+) = 100, 000 amp/sec 2 dt 2 Q#5.2: In the given network, K is closed at t = 0 with zero current in the inductor. Find the values of i, di/dt, and d 2 i/dt 2 at t = 0+ if R 10 L 1 H V 100 V K + R V  L www.electrical.net.tc 2010 www.electrical.net.tc Page 3 Key closed at t = 0 i L (0+) = i L (0) = i(0+) = 0 Amp According to the statement under Q#5.2: Drop Rise Open circuit Drop i(0+) Element and initial condition Equivalent circuit at t = 0+ oc Applying KVL for t 0 Sum of voltage rise = sum of voltage drop Ldi V = iR + dt Ldi = V – iR dt di V  iR = [eq. 1] dt L di(0+) V – i(0+)R = dt L Putting corresponding values di(0+) V – (0)R = dt L di(0+) V = dt L di(0+) 100 = dt 1 di(0+) = 100 Amp/sec dt www.electrical.net.tc 2010 www.electrical.net.tc Page 4 Differentiating [eq. 1] d 2 i d V iR =  dt 2 dt L L d 2 i Rdi = dt 2 Ldt d 2 i(0+) Rdi(0+) = dt 2 Ldt Putting corresponding values d 2 i(0+) = 1, 000 Amp/sec 2 dt 2 Q#5.3: In the network of the figure, K is changed from position a to b at t = 0. Solve for i, di/dt, and d 2 i/dt 2 at t = 0+ if R 1000 L 1 H C 0.1 F V 100 V a K b R V C L Equivalent circuit at t = 0+ b www.electrical.net.tc 2010 www.electrical.net.tc Page 5 sc Applying KVL for t 0 Sum of voltage rise = sum of voltage drop 1 Ldi Ri + idt + = 0 [eq. 1] C dt Equivalent circuit at t = 0 a i(0+) sc V 100 i L (0+) = i L (0) = i(0+) = = = 0.1 Amp R 1000 Initial condition: V C (0) = V C (0+) = 0 Also we know for t 0 V R + V L + V C = 0 iR + V L + V C = 0 At t = 0+ i(0+)R + V L (0+) + V C (0+) = 0 (0.1)(1000) + V L (0+) + 0 = 0 V L (0+) = 100 Volts And di V L = L dt di V L = dt L di(0+) V L (0+) = dt L Putting corresponding values di(0+) www.electrical.net.tc 2010 www.electrical.net.tc Page 6 = 100 Amp/sec dt Differentiating [eq. 1] with respect to „t‟ Rdi i Ld 2 i + + = 0 dt C dt 2 Rdi(0+) i(0+) Ld 2 i(0+) + + = 0 dt C dt 2 Putting corresponding values...
View
Full
Document
This note was uploaded on 10/07/2011 for the course EE 10 taught by Professor Chang during the Spring '07 term at UCLA.
 Spring '07
 Chang

Click to edit the document details