Test: Inductor Filters & Capacitor Filters - UPSC MCQ

Presence of inductor usually dampens the AC signal. Due to self induction induces opposing EMF or changes in the current.

Ripple factor will decrease when L is increased and R_L. Inductor has a higher dc resistance. It depends on property of opposing the change of direction of current.

 The inductor does not allow the ac components to pass through the filter. The main purpose of using an inductor filter is to avoid the ripples. By using this property, the inductor offers an infinite resistance to ac components and gives a smooth output.

For a rectifier with an inductor filter,
V_DC=2V_m/π, I_dc=V_DC/R_L=2V_m/R_Lπ
I_DC=2*250/(3.14*15*103)=10.6mA.

For any electronic devices, a steady dc output is required. The filter is used for this purpose. The ac components are removed by using a filter.

 Given the ripple voltage is 7V. So, 7=1.414V_RMS
ϒ=V_RMS/V_DC=4.95/380=0.0130. ϒ=1/3√2(R_L/L_ω)
So, L=28.8henry.

When R_Lis infinite, the ripple factor is 0.471. This value is close to that of a rectifier. So, the resistance should be small.

The inductor with high resistance can cause poor voltage regulation. The choke resistance, the resistance of half of transformer secondary is not negligible.

When the output current of a rectifier increases above a certain value, magnetic energy is stored in the inductor. This energy tends to decrease the sudden rise in the current. This also helps to prevent the current to fall down too much.

 ϒ=I_AC/I_DC, I_AC=2√2V_m/3π(R_L²+4ω²_L²)^1/2
By putting the values, I_AC=4.24Ma. V_DC=2V_m/π, I_DC=V_DC/R_L=2V_m/R_L π
I_DC=2*250/(3.14*15*10³)=10.6mA. ϒ=4.24/10.6=0.4.

Filtering is frequently done by shunting the load with capacitor. It depends on the fact that a capacitor stores energy when conducting and delivers energy during non conduction. Throughout this process, the ripples are eliminated.

The capacitor charges when the diode is in ON state and discharges during the OFF state of the diode. The instant at which the conduction starts is called cut-in point. The instant at which the conduction stops is called cut-out point.

 The diode permits charge to flow in capacitor when the transformer voltage exceeds the capacitor voltage. It disconnects the power source when the transformer voltage falls below that of a capacitor.

For a half wave filter,
ϒ=1/2√3 fCR_L=1/2√3*50*10^-3*R_L
R_L=10³/5√3=115.47Ω.

For a full wave rectifier, ϒ=1/4√3 fCR_L
=1/4√3*50*10^-3*2.5
=0.01154. So, ripple is 1.154%.

The ripple factor ϒ=I_L/ 4√3 fCV_DC
V_DC=200*10^-3/ 4√3 *50*500*8
=14.43.

The ‘T’ is the total non conducting time of capacitor. The charge per unit time will give the current flow.

The rectifier may be full wave or half wave. The capacitors are usually electrolytic even though they are large in size.

The ripple waveform will be triangular in nature. The rms value of this wave is independent of slopes or lengths of straight lines. It depends only on the peak value.

We know, V_m=V_dc+I_dc/4fC
=14.43+ {200/ (200*500)} 10³
=14.43+2=16.43V.