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Chapter 14-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

This is solution to problems related Electrical Circuit Analysis course. It was given by Prof. Gurnam Kanth at Punjab Engineering College. Its main points are: Transfer, Function, RC, circuit, Frequency, Response, Power, Average, Current, Division, Inductor, Capacitor

Typology: Exercises

2011/2012

Uploaded on 07/20/2012

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Download Chapter 14-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 14, Problem 1. Find the transfer function V o /V i of the RC circuit in Fig. 14.68. Express it using oω = 1/RC. Figure 14.68 For Prob. 14.1. Chapter 14, Solution 1. RCj1 RCj Cj1R R )( i o ω+ ω = ω+ ==ω V V H =ω)(H 0 0 j1 j ωω+ ωω , where RC 1 0 =ω 2 0 0 )(1 )(H ωω+ ωω =ω= H ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ω ω − π =ω∠=φ 0 1-tan 2 )(H This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that RC10 =ω . Thus, the sketches of H and φ are shown below. 0 90° 45° ω0 = 1/RC ω φ 0 1 0.7071 ω0 = 1/RC ω H docsity.com Chapter 14, Problem 2. Obtain the transfer function V o (s)/V i of the circuit in Fig. 14.69. Figure 14.69 For Prob. 14.2. Chapter 14, Solution 2. 6667.0s 4s 6 1 s/812 s/82 8/s 12010 8/s 12 V V )s(H i o + + = + + = ++ + == docsity.com Chapter 14, Problem 5. For each of the circuits shown in Fig. 14.72, find H(s) = V o (s)/V s (s). Figure 14.72 For Prob. 14.5. Chapter 14, Solution 5. (a) Let // sRLZ R sL R sL = = + o s s ZV V Z R = + ( ) ( ) o s s s s s sRL V Z sRLR sLH s sRLV Z R RR s R R LR R sL += = = = + + ++ + (b) Let 1 1// 1 1 Rx RsCZ R sC sRCR sC = = = ++ o s ZV V Z sL = + RsLLRCs R sRC1 RsL sRC1 R sLZ Z V V )s(H 2i o ++ = + + += + == docsity.com Chapter 14, Problem 6. For the circuit shown in Fig. 14.73, find H(s) = I o (s)/I s (s). Figure 14.73 For Prob. 14.6. Chapter 14, Solution 6. 1H j L sL sω⎯⎯→ = = Let //1 1 sZ s s = = + We convert the current source to a voltage source as shown below. 2 2 1( 1) 1 ( 1) 3 11 1 s s o s s s sI sIZ sV I x IsZ s s s s ss s += = = = + + + + + ++ + + 21 3 1 o s o V sII s s = = + + 2( ) 3 1 o s I sH s I s s = = + + + _ Vo Is ⋅ 1 Z 1 S + _ docsity.com Chapter 14, Problem 7. Calculate )(ωH if H dB equals (a) 0.05dB (b) -6.2 dB (c) 104.7 dB Chapter 14, Solution 7. (a) Hlog2005.0 10= Hlog105.2 10-3 =× == × -3105.210H 005773.1 (b) Hlog206.2- 10= Hlog0.31- 10= == -0.3110H 4898.0 (c) Hlog207.104 10= Hlog235.5 10= == 235.510H 510718.1 × Chapter 14, Problem 8. Determine the magnitude (in dB) and the phase (in degrees) of H(ω ) = at ω = 1 if H ( )ω equals (a) 0.05 dB (b) 125 (c) ω ω j j +2 10 (d) ωj+1 3 + ωj+2 6 Chapter 14, Solution 8. (a) 05.0H = == 05.0log20H 10dB 26.02- , =φ °0 (b) 125H = == 125log20H 10dB 41.94 , =φ °0 (c) °∠= + = 43.63472.4 j2 10j )1(H == 472.4log20H 10dB 01.13 , =φ °43.63 (d) °∠=−= + + + = 34.7-743.47.2j9.3 j2 6 j1 3)1(H = , =φ –34.7˚ docsity.com ω ) = )2( 10 ωω ω jj j + + Chapter 14, Solution 11. )2j1(j )10j1(5 )( ω+ω ω+ =ωH 2j1log20jlog2010j1log205log20H 10101010dB ω+−ω−ω++= 2tan10tan90- -1-1 ω−ω+°=φ The magnitude and phase plots are shown below. φ -45° 45° ω1001 10 0.1 -90° 90° HdB -20 20 ω1001 10 0.1 -40 40 34 14 docsity.com )10( 1 + + ss s Sketch the magnitude and phase Bode plots. docsity.com Chapter 14, Solution 12. 201.0log20, )10/1( )1(1.0)( −= + + = ωω ω jj jwT The plots are shown below. |T| (db) 20 0 ω 0.1 1 10 100 -20 -40 arg T 90o 0 ω 0.1 1 10 100 -90o docsity.com