Part b single phase thyristor full bridge rectifier nominal...

Part B: Single-Phase Thyristor Full Bridge Rectifier, Nominal values: Vs(rms) = 120V at 60Hz L = 100mH, R = 5Ω, delay angle α = 45° Case 1: without freewheeling diode Case 2: with freewheeling diode For each case, use MATLAB Simulink to implement the circuit and do the following: (a) Plot the current through T2 (b) Plot the voltage across T2 (c) Plot the output voltage v0 (d) Plot the source current is (e) Determine the peak and the average current in T1 (f) Determine the rms value of the line current (g) Determine the real input power (Watts) (h) Determine the VA apparent power (i) Determine the output power factor (j) Determine the input power factor Can you please solve case number 2?

Transcript

00:01 So, here in this question we are considering about the full wave bridge rectifier.

00:08 So, we are considering about the full wave bridge rectifier which is given here in the question.

00:14 In the first part we have to find out its efficiency, form factor, ripple factor, peak inverse voltage, transformation, universe from here.

00:22 So if we so here we are considering the value of the load is given that is equal to 100 ohm resistance is here.

00:29 So, in the first part we have to find out its efficiency that is neta that is equal to p of average divided by p of rms which is multiplied by the 100.

00:39 So, we are having this values here.

00:40 So, this is equal to 116 .698 which is divided by 144 .02.

00:46 So, we get the value of neta that become equals to 81 .028.

00:51 This is the value of the efficiency hence the answer to the first part.

00:54 In the second part we have to find out the form factor and form factor is equal to v rms which is divided by v average that from here is equal to 120 .017 which is divided by 108 .054.

01:10 So, the value of the form factor that is represented by ff become equals to 1 .11.

01:16 Hence the answer to the first part.

01:18 Second part now in the third part we have to find out the value of the ripple factor that is square root of irms divided by the i average minus 1.

01:28 So, this is the value of the ripple factor plug into the value that become equals to 1 .2 divided by 1 .08 to its whole square minus 1.

01:36 So, the value of rf from here is equals to 0 .4843.

01:41 This is the value of the ripple factor.

01:43 Hence the answer to the third part.

01:45 In the fourth part we have to find out the value of the peak inverse voltage.

01:50 So, the value of peak inverse voltage from here is represented by v m that is equals to 169 .7 that is involved.

01:58 This is the value of the v m from here.

02:00 Hence the answer to the fourth part.

02:02 Now in the fifth part we have to find out the value of tuf that is equals to p of dc which is divided by p of ac that is rated from here which from here is equals to lossless minus condition lossless minus condition minus assumed...

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