Power Electronics Test 5

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Power Electronics Test 5

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Weekly Quiz Competition

Question 1
1 / -0

A 3-phase full converter feeds power to an R load of 10 Ω. For a firing angle delay of 30° the load takes 5 kW. An inductor of large value is also connected ‘to’ the load to make current ripple free. The value of per phase input voltage is ________ (in V)

A
110-110.6

B
1104-110.61

C
1107-110.66

D
1109-110.62

Solution

Concept:
Average output voltage, \({V_0} = \frac{{3\;{V_{mL}}}}{\pi }\cos \alpha \)
For constant load current,
RMS value of load current = average load current
⇒ I_0r = I₀
Load power, \({P_0} = \frac{{V_0^2}}{R}\)
Calculation:
Given that, load power = 5 kW
Resistance (R) = 10 Ω
\( \Rightarrow \frac{{V_0^2}}{{10}} = 5 \times {10^3}\)
⇒ V₀ = 223.606 V
\( \Rightarrow \frac{{3 \times {V_{ph}} \times \sqrt 3 \times \sqrt 2 }}{\pi }\cos 30^\circ = 223\)
⇒ V_Ph = 110.38 V
Question 2
1 / -0

A 3-phase full converter delivers a ripple free load current of 10 A with a firing angle delay of 45°. The input voltage is 3-phase, 400 V, 50 Hz. The value of 2^nd harmonic source current amplitude is _______

A
11.25 A

B
3.89 A

C
2.69 sin (ωt – 45°) A

D
0 A

Solution

In a three-phase full converter, the n^th harmonic component of source current is
\({I_{sn}} = \frac{{4\;{I_0}}}{{\sqrt 2 \;n\pi }}\sin \frac{{n\pi }}{3}\) for n = 1, 3, 5,…
Even harmonics are absent in a three-phase full converter.
Therefore, the value of the second harmonic source current is zero.
Question 3
1 / -0

A three-phase uncontrolled diode rectifier supplies a constant load current of 10 A and its supply voltage is 400 V line to line. The supply fundamental rms current is ______

A
10 A

B
5.77 A

C
8.16 A

D
7.79 A

Solution

Concept:
In a three-phase uncontrolled diode rectifier
Average output voltage, \({{V}_{o}}=\frac{3{{V}_{ml}}}{\pi }\)
Supply rms current \({{i}_{sr}}={{I}_{o}}\sqrt{\frac{2}{3}}\)
supply fundamental rms current
\({{i}_{s1}}=\frac{4{{I}_{o}}}{\pi }\sin \left( \frac{\pi }{3} \right)\sin \left( \omega t-\alpha \right)\)
AC power (or) supply power \(=\sqrt{3}~{{V}_{sr}}{{i}_{sr}}\)
\(=\sqrt{3}~{{V}_{sr}}{{I}_{o}}~\sqrt{\frac{2}{3}}\)
DC power (or) load power = V_oI_o
Calculation:
Given that, supply voltage (V_s) = 400 V
Load current (I_o) = 10 A
Supply fundamental rms current
\({{i}_{slr}}=\frac{2\sqrt{2}}{\pi }\times 10\times \frac{\sqrt{3}}{2}=7.79~A\)
Question 4
1 / -0

A 3-phase full bridge diode rectifier is fed from delta-star transformer. The voltage fed to the rectifier has a maximum value of V₂ from line to neutral so that the voltage of phase ‘a’ is V_a = V_m sin ωt. The phase sequence is a-b-c. At the instant the phase ‘a’ voltage is passing through zero, what will be the rectifier output voltage?

A
Zero

B
V_m

C
\(\sqrt 2 \;{V_m}\)

D
\(\sqrt 3 \;{V_m}\)

Solution

For star side maximum phase voltage = V_m
For a 3 – phase full bridge diode rectifier, output voltage varies between \(\sqrt 3 \;{V_m}\) and 1.5 V_m
At ωt = 0, rectifier output voltage = \(\sqrt 3 \;{V_m}\)
Question 5
1 / -0

A step-down delta-star transformer, with per-phase turns ratio of 5 is fed from a 3-phase 1100 V, 50 Hz source. The secondary of this transformer is connected through a 3-pulse type rectifier, which is feeding on R load. The average value of output voltage is _______ (in V)

A
257-257.5

B
2574-257.51

C
2577-257.56

D
2579-257.52

Solution

Concept:
The average output voltage of a 3-phase 3-pulse rectifier,
\({V_0} = \frac{{3{V_{mL}}}}{{2\pi }}\)
The average output voltage of a 3-phase 6-pulse rectifier,
\({V_0} = \frac{{3{V_{mL}}}}{\pi }\)
Calculation:
Given that,
Supply voltage of a step-down transformer = 1100 V
Primary line voltage (V_P-L) = 1100 V
Primary phase voltage (V_P-Ph) = V_PL = 1100 V
Per-phase turns ratio = 5
Secondary phase voltage \(\left( {{V_{S - ph}}} \right) = \frac{{1100}}{5} = 220\;V\)
Secondary line voltage \(\left( {{V_{S - L}}} \right) = 220 \times \sqrt 3 \;V\)
Input supply of a 3-pulse rectifier \( = 220\sqrt 3 \;V\)
Average output voltage \(\left( {{V_0}} \right) = \frac{{3 \times 220 \times \sqrt 3 \times \sqrt 2 }}{{2\pi }}\)
= 257.3 V
Question 6
1 / -0

A three-phase fully controlled bridge converter is supplied from 400 V, 50 Hz, three-phase mains and operates at a firing angle 45°. If the current on load side is constant at 8 A and the load voltage is 320 V, the value of source inductance is _______ (in mH)

A
25.5-26.1

B
25.54-26.11

C
25.57-26.16

D
25.59-26.12

Solution

Concept:
The average output voltage of a three-phase fully controlled bridge converter by considering the effect of source inductance is
\({V_0} = \frac{{3\;{V_{mL}}}}{\pi }\cos \alpha - \frac{{3\;\omega {L_s}}}{\pi }{I_0}\)
Calculation:
Given that,
Load current (I₀) = 8 A
Firing angle (α) = 45°
Supply line voltage (V_L) = 400 V
Output voltage (V₀) = 320 V
Frequency (f) = 50 Hz
\( \Rightarrow 320 = \frac{{3 \times \sqrt 2 \times 400}}{\pi } \times \cos 45^\circ - \frac{{3 \times 2\pi \times 50 \times {L_s}}}{\pi } \times 8\)
⇒ L_s = 25.82 mH
Question 7
1 / -0

A 200 V, 1450 RPM, 100 A separately excited dc machine has an armature resistance of 0.04 Ω. The machine is driven by a three-phase half-controlled converter operating from a three phase 220 V, 50 Hz supply. The motor operates at the rated speed and rated torque. Assuming continuous conduction. The input distortion factor is _______

A
0.8-0.82

B
0.84-0.821

C
0.87-0.826

D
0.89-0.822

Solution

Concept:
In a three-phase half-controlled converter,
The average output voltage is, \({V_0} = \frac{{3{V_{ml}}}}{{2\pi }}\left( {1 + \cos \alpha } \right)\)
The distortion factor \( = \sqrt {\frac{6}{{\pi \left( {\pi - \alpha } \right)}}} \cos \frac{\alpha }{2}\)
Calculation:
Given that, supply voltage (V_L) = 220 V
Average output voltage (V₀) = 200 V
\( \Rightarrow 200 = \frac{{3 \times 220 \times \sqrt 2 }}{{2\pi }}\left( {1 + \cos \alpha } \right)\)
⇒ α = 69.73°
Input distortion factor is \( = \sqrt {\frac{6}{{\pi \left( {\pi - \alpha } \right)}}} \cos \frac{\alpha }{2} = 0.812\)
Question 8
1 / -0

A three-phase half wave rectifier supplied by a 220 V(rms) at 50 Hz and the load is of 1 kW at 200 V. purely resistive. Power consumed by the load with given supply voltage is _______ (in W)

A
560-580

B
5604-5801

C
5607-5806

D
5609-5802

Solution

Concept:
In a three-phase half wave rectifier
RMS voltage, \({{V}_{rms}}={{V}_{mp}}\sqrt{\frac{1}{2}+\frac{2\sqrt{3}}{8\pi }}=0.84~{{V}_{mp}}\)
Average voltage \({{V}_{avg}}=\frac{3\sqrt{3}}{2\pi }{{V}_{mp}}\)
V_mp is the maximum phase voltage
Calculation:
Given that,
Supply rms voltage = 220 V
Maximum phase voltage \({{V}_{mp}}=\frac{220\times \sqrt{2}}{\sqrt{3}}\)
= 179.63 V
V_rms= 0.84 × 179.63 = 150.88 V
Load voltage (V_o) = 200 V
Load is purely resistive \({{P}_{o}}=\frac{V_{o}^{2}}{R}\)
\(\Rightarrow R=\frac{{{\left( 200 \right)}^{2}}}{1\times {{10}^{3}}}=40~\text{ }\!\!\Omega\!\!\text{ }\)
Power consumed by load with the supply voltage.
\(=\frac{V_{rms}^{2}}{R}=\frac{{{\left( 150.88 \right)}^{2}}}{40}=569.184~W\)
Question 9
1 / -0

A three-phase fully controlled bridge converter is connected to a 415 V supply, having a source resistance of 0.3 Ω and inductance of 1.2 mH per phase. The converter is working in the inversion mode at a firing advance angle of 30°. What is the average generator voltage for the conditions: dc current I_d = 60 A, thyristor drop = 1.5 V and f = 50 Hz?

A
180 V

B
545 V

C
230 V

D
240 V

Solution

For three phase fully controlled bridge converter in inversion mode,
\(\frac{{3{V_{ml}}}}{\pi }\cos \alpha = - E + {I_0}R\)
After including voltage drop of source resistance, inductance and voltage drop of SCR, the above equation becomes
\(\frac{{3{V_{ml}}}}{\pi }\cos \alpha - \frac{{3\omega {L_s}}}{\pi }{I_0} - 2{V_{SCR}} - 2{I_0}{R_S} = - E + {I_0}R\)
\(\Rightarrow \frac{{3 \times \sqrt 2 \times 415}}{\pi }\cos \left( {180 - 30} \right) - \frac{{3 \times 100\pi \times 1.2 \times {{10}^{ - 3}}}}{\pi } - 60 + 2 \times 60 \times 0.3 - 2 \times 1.5 = - E\)
⇒ E = 545.96 V
Question 10
1 / -0

A fully controlled three phase bridge is operating in inverting mode from an ac supply of 220 V, 50 Hz. The source inductance per phase is 0.32 mH. If the converter is operating with an angle of firing advance of 30°, calculate the maximum current that can be commuted, allowing a recovery angle (or) overlapping angle of 5°.

A
62.3 A

B
74.5 A

C
42.1 A

D
36.78 A

Solution

Concept:
For a 3-phase fully controlled bridge,
\({{I}_{0}}=\frac{{{V}_{ml}}}{2\omega {{L}_{s}}}[\cos \alpha -\cos \left( \alpha +\mu \right)]\)
V_ml is maximum line voltage
L_s is source inductance
α is conduct firing angle
μ is overlapping angle
Calculation:
Given that, V_L = 220 V
V_mL= \(220\sqrt{2}~V\)
L = 0.32 mH
F = 50 Hz
Firing advance angle = 30°
⇒ α = 180 – 30 = 150°
Overlapping angle (μ) = 5°
\({{I}_{0}}=\frac{220\times \sqrt{2}}{2\times 2\pi \times 50\times 0.32\times {{10}^{-3}}}[\cos 150-\cos \left( 150+5 \right)]~\)
= 62.33 A

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Re-Attempt Weekly Quiz Competition

Power Electronics Test 5

Result

Power Electronics Test 5

Score

-

Rank

-

SHARING IS CARING

Question 1

110-110.6

1104-110.61

1107-110.66

1109-110.62

Solution

Question 2

11.25 A

3.89 A

2.69 sin (ωt – 45°) A

0 A

Solution

Question 3

10 A

5.77 A

8.16 A

7.79 A

Solution

Question 4

Zero

Vm

\(\sqrt 2 \;{V_m}\)

\(\sqrt 3 \;{V_m}\)

Solution

Question 5

257-257.5

2574-257.51

2577-257.56

2579-257.52

Solution

Question 6

25.5-26.1

25.54-26.11

25.57-26.16

25.59-26.12

Solution

Question 7

0.8-0.82

0.84-0.821

0.87-0.826

0.89-0.822

Solution

Question 8

560-580

5604-5801

5607-5806

5609-5802

Solution

Question 9

180 V

545 V

230 V

240 V

Solution

Question 10

62.3 A

74.5 A

42.1 A

36.78 A

Solution

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