Test: Filters - Electrical Engineering (EE) MCQ

The output contents along with the DC components the AC harmonics which does no useful work & reduces the efficiency.

 A rectifier is used to convert AC to DC. Lower the AC (Non-DC) components in the output lower are the ohmic losses.

Inductor (L) has a very important property that the current through it cannot change rapidly. We can make use of this property by connecting it in series.

 It offers high impedance to AC such as the AC ripples do not pass through the load.

 AC ripples are not allowed to pass, by S.C the AC ripples as the C is always connected in parallel with the load.

AC ripples are not allowed to pass, by S.C the AC ripples as the C is always connected in parallel with the load.

When the two are reversed biased and the other two SCR’s are not yet gated, the stored energy in the capacitor is supply the load R.

It is to be noted that the waveform is that of a voltage parameter. The voltage goes up for a very small time and than goes down (discharges) for longer time. For every half wave the peak of the waveform is to the left, if it were to be on the right it would be more likely to be a L filter.

When the load current is increasing, the capacitor is charging through the source voltage & its current keeps on falling & voltage rises.

Use,
C = K * [1 + 1/√2RF] Where, K = 1/4fR.

 C filter improves the efficiency and eliminates ripples by keeping the output voltage constant.

 Filter is always connected to the load side (output). Inductor has the property of keeping the current smooth and constant.

: I = Vo/R
Vo = 2Vm/π.

CRF = Ir/Io.

 When the load is reduced the resistance R is increased & ripple current increases.

The time constant RC is increased therefore the ripple factor is reduced.

The above waveform is that of the output current. The peak per half cycle is on the right, hence it is more likely to be an L filter.

 In case of L filter, if R is lowered, the time constant (L/R) increases, therefore ripple factor reduces. In case of a C filter, the time constant RC is increased, therefore, ripple factor is reduced.

Simply, an LC filter combines both the advantages of L and C filters.

The output voltage is given by,
Vo = (2Vm/π) – (4Vm/3π)cos 2ωt – (4Vm/15π)cos 4ωt . . . .
Hence the other values after the second factor (4Vm/3π) become very very small hence they can be eliminated.