49029099-Chap-5-nwa

49029099-Chap-5-nwa - www.electrical.net.tc 2010...

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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...
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This note was uploaded on 10/07/2011 for the course EE 10 taught by Professor Chang during the Spring '07 term at UCLA.

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49029099-Chap-5-nwa - www.electrical.net.tc 2010...

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