Chapter 09 Part 2-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

Download Chapter 09 Part 2-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 9, Problem 15. Evaluate these determinants: (a) j jj +−− −+ 15 32610 (b) °∠°∠ °−∠−°−∠ 453016 1043020 (c) jj jj jj + − −− 11 1 01 Chapter 9, Solution 15. (a) j1-5- 3j26j10 + −+ = -10 – j6 + j10 – 6 + 10 – j15 = –6 – j11 (b) °∠°∠ °∠°−∠ 453016 10-4-3020 = 60∠15° + 64∠-10° = 57.96 + j15.529 + 63.03 – j11.114 = 120.99 + j4.415 (c) j1j 0jj1 j1j1 j1j 0jj1 − −− + − −− = )j1(j)j1(j01011 22 ++−+−−++ = )j1j1(11 ++−− = 1 – 2 = –1 docsity.com Chapter 9, Problem 16. Transform the following sinusoids to phasors: (a) -10 cos (4t + 75 o ) (b) 5 sin(20t - 10 o ) (c) 4 cos2t + 3 sin 2t Chapter 9, Solution 16. (a) -10 cos(4t + 75°) = 10 cos(4t + 75° − 180°) = 10 cos(4t − 105°) The phasor form is 10∠-105° (b) 5 sin(20t – 10°) = 5 cos(20t – 10° – 90°) = 5 cos(20t – 100°) The phasor form is 5∠-100° (c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°) The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87° Chapter 9, Problem 17. Two voltages v1 and v2 appear in series so that their sum is v = v1 + v2. If v1 = 10 cos(50t - 3 π )V and v2 = 12cos(50t + 30 o ) V, find v. Chapter 9, Solution 17. 1 2 10 60 12 30 5 8.66 10.392 6 15.62 9.805 o o oV V V j j= + = < − + < = − + + = < − 15.62cos(50 9.805 ) Vov t= − = 15.62cos(50t–9.8˚) V docsity.com Chapter 9, Problem 21. Simplify the following: (a) f(t) = 5 cos(2t + 15(º) – 4sin(2t -30º) (b) g(t) = 8 sint + 4 cos(t + 50º) (c) h(t) = ∫ + t dttt 0 )40sin5040cos10( Chapter 9, Solution 21. (a) oooo jF 86.343236.8758.48296.690304155 ∠=+=−−∠−∠= )86.3430cos(324.8)( ottf += (b) ooo jG 49.62565.59358.4571.2504908 −∠=−=∠+−∠= )49.62cos(565.5)( ottg −= (c) ( ) 40,9050010 j 1H oo =ω−∠+∠ ω = i.e. ooo 69.1682748.125.125.0j18025.19025.0H −∠=−−=−∠+−∠= h(t) = 1.2748cos(40t – 168.69°) Chapter 9, Problem 22. An alternating voltage is given by v(t) = 20 cos(5t - 30 o ) V. Use phasors to find ∫ ∞− −+ t dttv dt dvtv )(24)(10 Assume that the value of the integral is zero at t = - ∞ . Chapter 9, Solution 22. Let f(t) = ∫ ∞− −+ t dttv dt dvtv )(24)(10 oV j VVjVF 3020,5,2410 −∠==−+= ω ω ω o89.334.454)10j32.17)(4.20j10(V4.0jV20jV10F ∠=−+=−+= )89.33t5cos(4.454)t(f o+= docsity.com Chapter 9, Problem 23. Apply phasor analysis to evaluate the following. (a) v = 50 cos(ω t + 30 o ) + 30 cos(ω t + 90 o )V (b) i = 15 cos(ω t + 45 o ) - 10 sin(ω t + 45 o )A Chapter 9, Solution 23. (a) 50 30 30 90 43.3 25 30 43.588 6.587o o oV j j= < + < = + − = < − 43.588cos( 6.587 ) Vov tω= − = 43.49cos(ωt–6.59˚) V (b) 15 45 10 45 90 (10.607 10.607) (7.071 7.071) 18.028 78.69o o o oI j j= < − < − = + − − = < 18.028cos( 78.69 ) Aoi tω= + = 18.028cos(ωt+78.69˚) A Chapter 9, Problem 24. Find v(t) in the following integrodifferential equations using the phasor approach: (a) v(t) + ∫ = tdtv cos10 (b) ∫ +=++ )104sin(204)(5 otdtvtvdt dv Chapter 9, Solution 24. (a) 1,010 j =ω°∠= ω + V V 10)j1( =−V °∠=+= − = 45071.75j5 j1 10 V Therefore, v(t) = 7.071 cos(t + 45°) (b) 4),9010(20 j 4 5j =ω°−°∠= ω ++ω V VV °∠=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ 80-20 4j 4 54jV docsity.com Chapter 9, Problem 25. Using phasors, determine i(t) in the following equations: (a) 2 )452cos(4)(3 otti dt di −=+ (b) 10 ∫ +=++ )225cos(5)(6 ottidt didti Chapter 9, Solution 25. (a) 2,45-432j =ω°∠=+ω II °∠=+ 45-4)4j3(I °∠= °∠ °∠ = + °∠ = 98.13-8.0 13.535 45-4 j43 45-4 I Therefore, i(t) = 0.8 cos(2t – 98.13°) (b) 5,2256j j 10 =ω°∠=+ω+ ω II I °∠=++ 225)65j2j-( I °∠= °∠ °∠ = + °∠ = 56.4-745.0 56.26708.6 225 3j6 225 I Therefore, i(t) = 0.745 cos(5t – 4.56°) Chapter 9, Problem 26. The loop equation for a series RLC circuit gives ∫ ∞− =++ t tdtii dt di 2cos2 Assuming that the value of the integral at t = -∞ is zero, find i(t) using the phasor method. Chapter 9, Solution 26. 2,01 j 2j =ω°∠= ω ++ω I II 1 2j 1 22j =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++I °∠= + = 87.36-4.0 5.1j2 1 I Therefore, i(t) = 0.4 cos(2t – 36.87°) docsity.com