Three Phase Full Converters MCQ Quiz in বাংলা - Objective Question with Answer for Three Phase Full Converters - বিনামূল্যে ডাউনলোড করুন [PDF]

Total rms current \({I_a} = \sqrt {\frac{2}{3}} \times 20 = 16.32\;A\)

Fundamental current \({I_{a1}} = 0.78 \times 20 = 15.6\;A\)

\(\begin{array}{l} THD = \sqrt {\frac{1}{{D{F^2}}} - 1} \\ DF = \frac{{{I_{{a_1}}}}}{{{I_a}}} = \frac{{0.78 \times 20}}{{0.816 \times 20}} = 0.955\\ THD = \sqrt {{{\left( {\frac{1}{{0.955}}} \right)}^2} - 1} = 30.73\% \end{array}\)

\({I_{0\left( {max} \right)}} = \frac{{{V_{0max}}}}{R} = \frac{{400}}{{10}} = 40A\)

Input transformer kVA rating = \(\sqrt 3 \times {V_l}{I_l}\)

 \({I_l}\) = Rms value of line current on ac diode \(= {I_0} \times \sqrt {\frac{2}{3}}\)

kVA rating of transformer \(= \sqrt 3 \times 400 \times 40 \times \sqrt {\frac{2}{3}}\)

= 22.6kVA

The Correct option is 3

Concept:

The output voltage of 3 phase 6 pulse bridge rectifiers is:

VO = \(\boldsymbol{\mathbf{}\frac{3V_{ML}}{\Pi }}\) = \(\frac{3\sqrt{3} V_{MP}}{\Pi }\)

Rms Line voltage = 415 V

Out voltage = V_O

Calculation:

Pick inverse voltage 3 phase 6 pulse bridge rectifiers if the safety factor is 2 

V _peak =   \(\ {\sqrt{2}\times saftey factor\times V_{rms} = \sqrt{2}\times 2.1\times 415}\) = 1232.49

Peak inverse voltage =V_peak

Additional Information Thyristor:

• A thyristor is a four-layer semiconductor device, consisting of alternating P-type and N-type materials (PNPN). A thyristor usually has three electrodes: an anode, a cathode, and a gate, also known as a control electrode.

Concept:

Ripple frequency of three-phase full-wave converter:

Figure: output voltage waveform of three-phase full-wave rectifier.

From the above output voltage waveform, we can observe that for a complete one cycle of input supply we got 6 pulses in the output.

So, the three-phase full-wave ac to dc converter is a 6-pulse converter.

Then the ripple frequency of the output f₀ = m × f

Where,

m = number of pulses in the output per one complete cycle of the input

f = supply voltage frequency

Solution:

For single-phase full-wave bridge circuit f0 = 2 × f

For a three-phase full-wave converter f₀ = 6 × f

Hence, the ratio output ripple-frequency to the supply-voltage frequency = f0 / f = 6

In a six-pulse thyristor bridge converter, the harmonics present in source or current are:

6k ± 1, k = 1, 2, 3 …

So, the harmonics present are 5, 7, 11, 13…

Harmonics frequencies are 250 Hz, 350 Hz, 550 Hz, 650 Hz…

Three-Phase Full Converters:

If all diode is replaced by thyristor, a three-phase full converter bridge formed as shown below,

• A three-phase input supply is connected to the terminal A, B, C, and RLE load is connected across the output terminal.
• It worked as three-phase AC to DC converter for firing angle delay 0 <α ≤ 90.
• The positive group of SCRs is fired at an interval of 120° and a similarly negative group of SCRs fired at an interval of 120°, but SCR of both groups are fired at an internal of 60° or commutation occurs at every 60°.
• At any time tow SCRs, one from the positive group and one from the negative group must be conducted together for the source to energize the load.
   

Waveform:

Voltage and Current Equation:

The average output voltage (V0) is given by,

\(V_0=\frac{3V_m}{\pi}cosα\ \longrightarrow 0≤ α≤ \pi\)

Where Vm is the maximum value of line voltage.

Voltage and current waveform for α = 30° and for constant current can be drawn as,

From waveform RMS value of Thyristor current (IT = iT1) will be,

\(I_T=\sqrt{I_0^2\times \frac{2\pi}{3}\times \frac{1}{2\pi}}=I_0\times \sqrt\frac{1}{3}\)

Where I0 is constant DC load current.

Assuming each secondary winding is identical and has the same resistance R

When the R phase has a maximum voltage, lines a and c will conduct. At this condition, the current will be divided into R and YB winding.

Current through \(R = \frac{{{I_0} \times 2R}}{{3R}} = \frac{{2{I_0}}}{3}\)

Current through B and Y = \(\frac{{{I_0}}}{3}\)

And \({I_0} = \frac{{2I}}{3}\)

Current through R and Y \( = \frac{{{I_0}}}{3}\)

Current through B \( = \frac{{2{I_0}}}{3}\)

The current through Δ-phase winding

Current through primary \({I_p} = k{I_s}\)

As the load is resistive the peak instantaneous output voltage is 

Vm = √2 Vrms

\({V_m} = 400√ 2 \;V\)

Given that:

There phase converter feeds purl resistive load.

Firing angle α = 60° 

There phase full converter is a fully controlled bridge converter using 6 thyristors connected in the form of a full wave bridge configuration

For firing angle α = 60°,

Now, average value of current is,

 \({I_{avg}} = \frac{{{V_{dc}}}}{R} = \frac{{3\sqrt 3 \;{V_m}}}{{\pi R}}\cos \alpha \)

For both R and RL load average current equation is same.

Therefore, its value is also same i.e. 10 A for R and RL load.