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Sir u have taught us that load commutation is class A commutation of forced comm. And this CSI C load is natural and load commutation both. How is that possible that something is natural as well as forced commutation both at the same time...sir plzz answer it..
Sir as u told , rms value for triangular wave=Ao/root(3), when we shift graph upward,then ampltude will be 2Vmax. Then Vrms=2V max/root(3), But ans is Vmax/ root(3).
17:09 this statement you may have to correct it out..since rms does change with shifting... this is basically triangular wave with dc offset...it s actual rms will be square root of the squares of the rms of triangular wave and the dc offset...In this case...Vrms(of sawtooth wave)= 2Ao/√3....DC offset rms=Ao.... Vin(rms)=√(Vrms^2 - DC offset^2) = Ao/√3 (desired answer )
Yes true, Irms=root(Iac2+Idc2) So, in the second part a dc counterpart is getting introduced and that would certainly change the rms of the new waveform
Sir here Vin is negative from 0 to T/4 and Is is constant as always..so power is going to be negative then power would flow from load to source??is it right??
@@VarunKumar-bc7vr since it's a capacitive load the voltage direction will not reverse quickly...so after π/2 only it will be reverse biased..hence circuit turn off time will be π/2
Why don't we take capacitive load in case of vsi? I can see that capacitive load on csi and inductive load in vsi are analogous to each other, may be reason lies in it
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Hats off sir😎😎😎 best teacher
Most welcome..all the best for gate
Nice teaching. thanks for uploading free videos
Nice video
Awesome
Sir u have taught us that load commutation is class A commutation of forced comm. And this CSI C load is natural and load commutation both. How is that possible that something is natural as well as forced commutation both at the same time...sir plzz answer it..
It is not natural commutation. Natural commutation is a characteristic property of the source. Here CSI C load is Class A or Self or Load commutation.
Sir as u told , rms value for triangular wave=Ao/root(3), when we shift graph upward,then ampltude will be 2Vmax. Then Vrms=2V max/root(3),
But ans is Vmax/ root(3).
Same doubt here
No, amplitude will not be 2Vmax.... correct yourself...and hence conclusion is "upward and downward shifts will not change RMS value"...
Even without shifting the graph upwards, and maintaining the original waveform,
Vrms=2Vmax/root(3).
17:09 this statement you may have to correct it out..since rms does change with shifting... this is basically triangular wave with dc offset...it s actual rms will be square root of the squares of the rms of triangular wave and the dc offset...In this case...Vrms(of sawtooth wave)= 2Ao/√3....DC offset rms=Ao....
Vin(rms)=√(Vrms^2 - DC offset^2) = Ao/√3 (desired answer )
Yes true, Irms=root(Iac2+Idc2)
So, in the second part a dc counterpart is getting introduced and that would certainly change the rms of the new waveform
Sir output current sinusoidal hoga ya square wave
Sir here Vin is negative from 0 to T/4 and Is is constant as always..so power is going to be negative then power would flow from load to source??is it right??
Right
thank you sir
@@GATEMATICEducation why circuit turn off time is pi/2 and not pi?
@@VarunKumar-bc7vr since it's a capacitive load the voltage direction will not reverse quickly...so after π/2 only it will be reverse biased..hence circuit turn off time will be π/2
sir you explain well but you can make it a bit short and concise as well.
Why don't we take capacitive load in case of vsi? I can see that capacitive load on csi and inductive load in vsi are analogous to each other, may be reason lies in it
Peak at 17:28 will become 2Vmax on shift then ans will not same
8:33 not natural commutation that is self commutation