laplace

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Explain the analysis of RLC network with and without initial condition with the help of laplace. 

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Trapezoidal rule 3

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Evaluate By using Trapezoidal rule Solution:- h= 10-0/10 = 1 y= f(x) = 1/ 1+ x^2 X            y 0            1 1            0.5 2            0.2 3            0.1 4            0.05882 5            0.03846 6            0.02702 7            0.02 8            0.01538 9            0.01219 10          0.00990 Using Trapezoidal rule Integration of Y dx = h/2 [ ( Y 0 + Y n) + 2( Y 1 + Y2+ .......... Y(n-1) ] => 1/2[ (1+0.00990) + 2(0.5+ 0.2 + 0.1 + 0.05882+0.03846 + 0.02702 + 0.02+ 0.01538 + 0.01219) ] => 0.5 [ 1.00990 + 2× 0.94485] => 0.5 [ 1.00990 + 1.88971 ] =>  1.449805
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Ripple factor and efficiency

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Derive the Ripple factor and efficiency in case of half wave and full wave rectifier circuit .  Ripple factor: - Ripple factor is the ratio of RMS value of AC component present in the rectified output to the average value of rectified output. it is  dimensionless quantity.  Rectification efficiency:- it is a quantitative measurement of rectifier efficiency . it is ratio of DC power output to AC power input to the rectifier. it is denoted and expressed in percentage.  Derivation of Ripple factor: By definition, The RMS value of total load current is given by I rms = √( I dc)^2+ ( I ac) ^2 (or) Iac = √ I rms ^2+ I dc^2 When the above equation is divided by using Idc then we can get the following equation. Iac / Idc = 1/ Idc √I2rms + I2dc However, here Iac / Idc is the ripple factor formula R.F = 1/ Idc √I2rms + I2dc = √ (Irms / Idc)2 -1 Ripple Factor of Half Wave Rectifier For half-wave rectifier, Irms = Im/2 Idc = Im/ π We know the formula of R.F = √ (Irms / Idc...
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Trapezoidal rule 6

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Evaluate Taking 4 equal intervals (I) Trapezoidal rule  And (ii) Simpsons 1/3 rd rule Solution:- h = 0.4 - 0 /4 = 0.1 Y = f(x) = cos x X            Y  0            1 0.1         0.99500 0.2         0.98006 0.3         0.95534 0.4         0.92106 (I) using Trapezoidal rule Integration of y dx = h/2[(y0+ yn) + 2( y1+ y2..... + y(n-1) ] => 0.1/2 [ (1+ 0.92106) + 2( 0.99500+ 0.98006 + 0.95534) ] => 0.1/2 [ 1.92106 + 2 × 2.9304 ] => 0.05 [ 1.92106 + 5.8608] => 0.05 × 7.78186  => 0.389093 (II)    using Simpsons 1/3rd rule Integration of y dx = h/3[(y0+ yn) + 4( y1+ y3 ..... + y(n-1) +2 ( y2+ y4 ..... + y(n-2)] => 0.1/3 [ (1+ 0.92106) + 4(0.99500+ 0.95534) + 2 × 0.98006 ] => 0.1/3 [ 1.92106 + 4× 1.95034 + 2× 0.98006 ] => 0.1/3 [ 1.92106 + 7.80136 + 1.96012]  => 0.1/3 ...
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