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Chapter 16-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: Laplace, Transform, S-domain, Circuit, Superposition, Transform, Voltage, Source, Function, Signal

Typology: Exercises

2011/2012

Uploaded on 07/20/2012

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Download Chapter 16-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 16, Problem 1. Determine i(t) in the circuit of Fig. 16.35 by means of the Laplace transform. Figure 16.35 For Prob. 16.1. Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 222 )23()21s( 1 1ss 1 s1s1 s1 )s(I ++ = ++ = ++ = โŽŸโŽŸ โŽ  โŽž โŽœโŽœ โŽ โŽ› = t 2 3sine 3 2)t(i 2t- =)t(i A)t866.0(sine155.1 -0.5t 1/s 1 + โˆ’ 1/s s I(s) docsity.com Chapter 16, Problem 2. Find v x in the circuit shown in Fig. 16.36 given v s .= 4u(t)V. Figure 16.36 For Prob. 16.2. docsity.com Chapter 16, Solution 3. In the s-domain, the circuit becomes that shown below. 1 I 4 2 s + 2 0.2s We transform the current source to a voltage source and obtain the circuit shown below. 2 1 I 8 4 s + 0.2s 8 4 20 40 3 0.2 ( 15) 15 s A BsI s s s s s + + = = = + + + + 40 8 15 20 40 52, 15 3 15 3 xA B โˆ’ += = = = โˆ’ 8 / 3 52 / 3 15 I s s = + + 158 52( ) ( ) 3 3 ti t e u tโˆ’โŽก โŽค= +โŽข โŽฅโŽฃ โŽฆ + _ docsity.com Figure 16.38 For Prob. 16.4. Chapter 16, Solution 4. The circuit in the s-domain is shown below. I 2 4I + 5 Vo 1/s 1 โ€“ 4 5 1/ o o VI I I sV s + = โŽฏโŽฏโ†’ = But 5 2 oVI โˆ’= 5 12.55 2 5 / 2 o o o V sV V s โˆ’โŽ› โŽž = โŽฏโŽฏโ†’ =โŽœ โŽŸ +โŽ โŽ  2.5( ) 12.5 Vtov t e โˆ’= + _ docsity.com Chapter 16, Problem 5. If i s (t) = e t2โˆ’ u(t) A in the circuit shown in Fig. 16.39, find the value of i 0 (t). Figure 16.39 For Prob. 16.5. Chapter 16, Solution 5. ( ) ( ) A)t(ut3229.1sin7559.0e or A)t(ueee3779.0eee3779.0e)t(i 3229.1j5.0s )646.2j)(3229.1j5.1( )3229.1j5.0( 3229.1j5.0s )646.2j)(3229.1j5.1( )3229.1j5.0( 2s 1 )3229.1j5.0s)(3229.1j5.0s)(2s( s 2 VsI )3229.1j5.0s)(3229.1j5.0s)(2s( s2 2ss s2 2s 1 2 s 2 1 s 1 1 2s 1V t2 t3229.1j2/t90t3229.1j2/t90t2 o 22 2 o 2 โˆ’= ++= โˆ’+ ++ +โˆ’ + ++ โˆ’โˆ’ โˆ’โˆ’ + + = โˆ’++++ == โˆ’++++ =โŽŸโŽŸ โŽ  โŽž โŽœโŽœ โŽ โŽ› +++ = โŽŸ โŽŸ โŽŸ โŽŸ โŽ  โŽž โŽœ โŽœ โŽœ โŽœ โŽ โŽ› +++ = โˆ’ โˆ’ยฐโˆ’โˆ’ยฐโˆ’โˆ’ or io(t) = A)t(ut2 7sin 7 2e t2 โŽŸ โŽŸ โŽ  โŽž โŽœ โŽœ โŽ โŽ› โŽŸโŽŸ โŽ  โŽž โŽœโŽœ โŽ โŽ› โˆ’โˆ’ s s 2 2s 1 + Io 2 docsity.com
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