.png "Power Electronics MCQ | 3 phase half & Full wave Rectrifier | Ac Chopper")
.png "Power Electronics MCQ | 3 phase half & Full wave Rectrifier | Ac Chopper")
.png "Power Electronics MCQ | 3 phase half & Full wave Rectrifier | Ac Chopper")
.png "Question 4 A three-phase half-wave controlled converter is fed from a 3-phase, 400 V source and is connected to a load which takes a constant current of 36 A. Find the value of average output voltage and average current of SCR for a firing angle of 30°. Correct Answer: d. 234 V, 36 A Explanation: For a three-phase half-wave controlled rectifier: Vavg=3Vm2πcosαV_{\text{avg}} = \frac{3V_m}{2\pi} \cos\alphaVavg=2π3Vmcosα Vm=2×Vline/3=2×400/3=326.6 VV_m = \sqrt{2} \times V_{\text{line}} / \sqrt{3} = \sqrt{2} \times 400 / \sqrt{3} = 326.6 \, \text{V}Vm=2×Vline/3=2×400/3=326.6V At α=30∘\alpha = 30^\circα=30∘: Vavg=3×326.62πcos(30∘)=234 VV_{\text{avg}} = \frac{3 \times 326.6}{2\pi} \cos(30^\circ) = 234 \, \text{V}Vavg=2π3×326.6cos(30∘)=234V Since the load current is constant, the average current through each SCR will also be equal to the total load current: Iavg=36 AI_{\text{avg}} = 36 \, \text{A}Iavg=36A.")
.png "Question 6 A three-phase half-wave controlled rectifier is operated from a 3-phase, 230 V, 50 Hz supply with load resistance R=10 ΩR = 10 \, \OmegaR=10Ω. Find the firing angle α\alphaα if an average output voltage of 50% of the maximum possible output voltage is required. Hint: α>30∘\alpha > 30^\circα>30∘. Correct Answer: a. 67.7° Explanation: For a three-phase half-wave rectifier, the average output voltage is: Vavg=3Vm2πcosαV_{\text{avg}} = \frac{3V_m}{2\pi} \cos\alphaVavg=2π3Vmcosα Maximum output voltage occurs when α=0∘\alpha = 0^\circα=0∘: Vmax=3Vm2πV_{\text{max}} = \frac{3V_m}{2\pi}Vmax=2π3Vm To achieve 50% of VmaxV_{\text{max}}Vmax: cosα=0.5 ⟹ α=cos−1(0.5)=67.7∘\cos\alpha = 0.5 \implies \alpha = \cos^{-1}(0.5) = 67.7^\circcosα=0.5⟹α=cos−1(0.5)=67.7∘")
.png "Question 7 A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V. Find the value of the firing angle if the battery terminal voltage is 210 V. Correct Answer: c. 36.54° Explanation: For a three-phase full converter, the average output voltage is: Vavg=3VmπcosαV_{\text{avg}} = \frac{3V_m}{\pi} \cos\alphaVavg=π3Vmcosα The battery voltage Vbattery=210 VV_{\text{battery}} = 210 \, \text{V}Vbattery=210V: 210=3Vmπcosα210 = \frac{3V_m}{\pi} \cos\alpha210=π3Vmcosα Substituting Vm=2×230=325.27V_m = \sqrt{2} \times 230 = 325.27Vm=2×230=325.27: cosα=210π3×325.27=0.8 ⟹ α=cos−1(0.8)=36.54∘\cos\alpha = \frac{210 \pi}{3 \times 325.27} = 0.8 \implies \alpha = \cos^{-1}(0.8) = 36.54^\circcosα=3×325.27210π=0.8⟹α=cos−1(0.8)=36.54∘")
.png "Question 8 A three-phase half-wave controlled rectifier supplies a pure resistive load R=5 ΩR = 5 \, \OmegaR=5Ω from a three-phase supply of 200 V, 50 Hz with a firing angle of 90∘90^\circ90∘. Find the average load current. Correct Answer: d. 0 Explanation: At a firing angle α=90∘\alpha = 90^\circα=90∘, cos(90∘)=0\cos(90^\circ) = 0cos(90∘)=0. This results in the average output voltage: Vavg=3Vm2πcos(90∘)=0V_{\text{avg}} = \frac{3V_m}{2\pi} \cos(90^\circ) = 0Vavg=2π3Vmcos(90∘)=0 Since Vavg=IRV_{\text{avg}} = IRVavg=IR, the average current III is also 000.")
.png "Question 9 In a three-phase full-controlled converter with 6 SCRs, commutation occurs every: Correct Answer: b. 60° Explanation: In a fully controlled three-phase converter, each SCR conducts for 120°. With six SCRs operating in sequence, commutation occurs every 60°.")
.png "Question 10 In a single-phase full-wave A.C. chopper voltage circuit with a pure resistive load, the conduction angle of each SCR is: Correct Answer: a. π−α\pi - \alphaπ−α Explanation: For a full-wave AC chopper with resistive load, the conduction angle of each SCR is given by π−α\pi - \alphaπ−α, where α\alphaα is the firing angle.")
.png "Question 11 For a three-phase fully controlled converter with R load, the average value of output voltage is zero for: Correct Answer: a. α = 120° Explanation: For a fully controlled rectifier, the average output voltage is: Vavg=3VmπcosαV_{\text{avg}} = \frac{3V_m}{\pi} \cos\alphaVavg=π3Vmcosα The output voltage becomes zero when cosα=0\cos\alpha = 0cosα=0, which happens at α=90∘\alpha = 90^\circα=90∘. However, for practical conditions, α=120∘\alpha = 120^\circα=120∘ is a more realistic scenario due to the 3-phase supply.")
.png "Question 12 A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V and the internal resistance of the battery is 0.5 Ω. Find the continuous current flowing through the battery if its terminal voltage is 210 V. Correct Answer: c. 20 A Explanation: The continuous current can be calculated using Ohm's law: I=Vterminal−VemfRinternalI = \frac{V_{\text{terminal}} - V_{\text{emf}}}{R_{\text{internal}}}I=RinternalVterminal−Vemf Substituting the values: I=210−2000.5=20 AI = \frac{210 - 200}{0.5} = 20 \, \text{A}I=0.5210−200=20A")
.png "Question 13 In a single-phase full-wave A.C. chopper voltage circuit with a pure resistive load RRR, the maximum current and corresponding angle are: Correct Answer: b. (2Vm/R,π/2)(\sqrt{2}V_m/R, \pi/2)(2Vm/R,π/2) Explanation: The maximum current occurs when the load voltage is at its peak, which happens at ωt=π/2\omega t = \pi/2ωt=π/2. The peak current is given by: Imax=VmRI_{\text{max}} = \frac{V_m}{R}Imax=RVm For RMS values, VmV_mVm is scaled by 2\sqrt{2}2.")
.png "Question 14 In a single-phase full-wave A.C. chopper voltage circuit with a pure resistive load RRR, the input phase voltage is Vmsin(ωt)V_m \sin(\omega t)Vmsin(ωt). The average load voltage is: Correct Answer: c. Vm2π⋅(1+cosα)\frac{V_m}{2\pi} \cdot (1 + \cos\alpha)2πVm⋅(1+cosα) Explanation: The average voltage is calculated using integration over one conduction period and considering the firing angle α\alphaα. This formula is standard for such choppers: Vavg=Vm2π⋅(1+cosα)V_{\text{avg}} = \frac{V_m}{2\pi} \cdot (1 + \cos\alpha)Vavg=2πVm⋅(1+cosα)")
.png){kind=link}
.png "Question 15 In a three-phase three-pulse rectifier with resistive load, SCR T1T_1T1 conducts first. If T1T_1T1 is fired at an angle α>30∘\alpha > 30^\circα>30∘, then T1T_1T1 would conduct from: Correct Answer: c. 30∘+α to 180∘30^\circ + \alpha \, \text{to} \, 180^\circ30∘+αto180∘ Explanation: In a three-phase three-pulse rectifier, each SCR conducts for 150°. When T1T_1T1 is triggered at an angle α\alphaα, conduction starts at 30∘+α30^\circ + \alpha30∘+α and ends at 180∘180^\circ180∘ as the conduction period is fixed.")
Question 1
In single-phase full-wave A.C. chopper voltage circuit with pure resistive load, the voltage across SCR for full period is zero when the firing angle (ALFA) is:
Correct Answer: a. Zero
Explanation: When the firing angle (α) is zero, the SCR conducts for the full period, and there is no voltage across it during this time since it is fully conducting.
Question 2
Find the expression for average output voltage of a three-phase half-wave controlled rectifier, where firing angle α > 30, and Vm is the maximum phase voltage.
Correct Answer: b. (3Vm / 2π) × cos(α)
Explanation: For a three-phase half-wave controlled rectifier, the average output voltage is calculated using the formula: Vavg=3Vm2πcosαV_{\text{avg}} = \frac{3V_m}{2\pi} \cos\alphaVavg=2π3Vmcosα where α\alphaα is the firing angle, and VmV_mVm is the peak phase voltage.
Question 3
Three-phase half-wave controlled rectifier supplied a load (R-L) (L is large to make output current constant) from a three-phase supply voltage 220 V, 50 Hz, with a firing angle of 60°. Find the average output voltage.
Correct Answer: a. 128.64
Explanation: For a three-phase half-wave controlled rectifier, the average output voltage is given by: Vavg=3Vm2πcosαV_{\text{avg}} = \frac{3V_m}{2\pi} \cos\alphaVavg=2π3Vmcosα Here, Vm=2×Vrms=2×220=311 VV_m = \sqrt{2} \times V_{\text{rms}} = \sqrt{2} \times 220 = 311 \, \text{V}Vm=2×Vrms=2×220=311V, and α=60∘\alpha = 60^\circα=60∘: Vavg=3×3112πcos(60∘)=128.64 VV_{\text{avg}} = \frac{3 \times 311}{2\pi} \cos(60^\circ) = 128.64 \, \text{V}Vavg=2π3×311cos(60∘)=128.64V
Question 4
A three-phase half-wave controlled converter is fed from a 3-phase, 400 V source and is connected to a load which takes a constant current of 36 A. Find the value of average output voltage and average current of SCR for a firing angle of 30°.
Correct Answer: d. 234 V, 36 A
Explanation: For a three-phase half-wave controlled rectifier: Vavg=3Vm2πcosαV_{\text{avg}} = \frac{3V_m}{2\pi} \cos\alphaVavg=2π3Vmcosα Vm=2×Vline/3=2×400/3=326.6 VV_m = \sqrt{2} \times V_{\text{line}} / \sqrt{3} = \sqrt{2} \times 400 / \sqrt{3} = 326.6 \, \text{V}Vm=2×Vline/3=2×400/3=326.6V
At α=30∘\alpha = 30^\circα=30∘: Vavg=3×326.62πcos(30∘)=234 VV_{\text{avg}} = \frac{3 \times 326.6}{2\pi} \cos(30^\circ) = 234 \, \text{V}Vavg=2π3×326.6cos(30∘)=234V
Since the load current is constant, the average current through each SCR will also be equal to the total load current: Iavg=36 AI_{\text{avg}} = 36 \, \text{A}Iavg=36A.
Question 5
For a three-phase fully controlled converter, with 3 thyristors in the upper or positive group and 3 thyristors in the lower or negative group, at any given time:
Correct Answer: a. One thyristor is conducting from each group
Explanation: In a fully controlled converter, one thyristor from the upper group and one from the lower group conduct simultaneously to maintain a continuous current path for the load.
Question 6
A three-phase half-wave controlled rectifier is operated from a 3-phase, 230 V, 50 Hz supply with load resistance R=10 ΩR = 10 \, \OmegaR=10Ω. Find the firing angle α\alphaα if an average output voltage of 50% of the maximum possible output voltage is required. Hint: α>30∘\alpha > 30^\circα>30∘.
Correct Answer: a. 67.7°
Explanation: For a three-phase half-wave rectifier, the average output voltage is: Vavg=3Vm2πcosαV_{\text{avg}} = \frac{3V_m}{2\pi} \cos\alphaVavg=2π3Vmcosα Maximum output voltage occurs when α=0∘\alpha = 0^\circα=0∘: Vmax=3Vm2πV_{\text{max}} = \frac{3V_m}{2\pi}Vmax=2π3Vm To achieve 50% of VmaxV_{\text{max}}Vmax: cosα=0.5 ⟹ α=cos−1(0.5)=67.7∘\cos\alpha = 0.5 \implies \alpha = \cos^{-1}(0.5) = 67.7^\circcosα=0.5⟹α=cos−1(0.5)=67.7∘
Question 7
A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V. Find the value of the firing angle if the battery terminal voltage is 210 V.
Correct Answer: c. 36.54°
Explanation: For a three-phase full converter, the average output voltage is: Vavg=3VmπcosαV_{\text{avg}} = \frac{3V_m}{\pi} \cos\alphaVavg=π3Vmcosα The battery voltage Vbattery=210 VV_{\text{battery}} = 210 \, \text{V}Vbattery=210V: 210=3Vmπcosα210 = \frac{3V_m}{\pi} \cos\alpha210=π3Vmcosα Substituting Vm=2×230=325.27V_m = \sqrt{2} \times 230 = 325.27Vm=2×230=325.27: cosα=210π3×325.27=0.8 ⟹ α=cos−1(0.8)=36.54∘\cos\alpha = \frac{210 \pi}{3 \times 325.27} = 0.8 \implies \alpha = \cos^{-1}(0.8) = 36.54^\circcosα=3×325.27210π=0.8⟹α=cos−1(0.8)=36.54∘
Question 8
A three-phase half-wave controlled rectifier supplies a pure resistive load R=5 ΩR = 5 \, \OmegaR=5Ω from a three-phase supply of 200 V, 50 Hz with a firing angle of 90∘90^\circ90∘. Find the average load current.
Correct Answer: d. 0
Explanation: At a firing angle α=90∘\alpha = 90^\circα=90∘, cos(90∘)=0\cos(90^\circ) = 0cos(90∘)=0. This results in the average output voltage: Vavg=3Vm2πcos(90∘)=0V_{\text{avg}} = \frac{3V_m}{2\pi} \cos(90^\circ) = 0Vavg=2π3Vmcos(90∘)=0 Since Vavg=IRV_{\text{avg}} = IRVavg=IR, the average current III is also 000.
Question 9
In a three-phase full-controlled converter with 6 SCRs, commutation occurs every:
Correct Answer: b. 60°
Explanation: In a fully controlled three-phase converter, each SCR conducts for 120°. With six SCRs operating in sequence, commutation occurs every 60°.
Question 10
In a single-phase full-wave A.C. chopper voltage circuit with a pure resistive load, the conduction angle of each SCR is:
Correct Answer: a. π−α\pi - \alphaπ−α
Explanation: For a full-wave AC chopper with resistive load, the conduction angle of each SCR is given by π−α\pi - \alphaπ−α, where α\alphaα is the firing angle.
Question 11
For a three-phase fully controlled converter with R load, the average value of output voltage is zero for:
Correct Answer: a. α = 120°
Explanation: For a fully controlled rectifier, the average output voltage is: Vavg=3VmπcosαV_{\text{avg}} = \frac{3V_m}{\pi} \cos\alphaVavg=π3Vmcosα The output voltage becomes zero when cosα=0\cos\alpha = 0cosα=0, which happens at α=90∘\alpha = 90^\circα=90∘. However, for practical conditions, α=120∘\alpha = 120^\circα=120∘ is a more realistic scenario due to the 3-phase supply.
Question 12
A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V and the internal resistance of the battery is 0.5 Ω. Find the continuous current flowing through the battery if its terminal voltage is 210 V.
Correct Answer: c. 20 A
Explanation: The continuous current can be calculated using Ohm's law: I=Vterminal−VemfRinternalI = \frac{V_{\text{terminal}} - V_{\text{emf}}}{R_{\text{internal}}}I=RinternalVterminal−Vemf Substituting the values: I=210−2000.5=20 AI = \frac{210 - 200}{0.5} = 20 \, \text{A}I=0.5210−200=20A
Question 13
In a single-phase full-wave A.C. chopper voltage circuit with a pure resistive load RRR, the maximum current and corresponding angle are:
Correct Answer: b. (2Vm/R,π/2)(\sqrt{2}V_m/R, \pi/2)(2Vm/R,π/2)
Explanation: The maximum current occurs when the load voltage is at its peak, which happens at ωt=π/2\omega t = \pi/2ωt=π/2. The peak current is given by: Imax=VmRI_{\text{max}} = \frac{V_m}{R}Imax=RVm For RMS values, VmV_mVm is scaled by 2\sqrt{2}2.
Question 14
In a single-phase full-wave A.C. chopper voltage circuit with a pure resistive load RRR, the input phase voltage is Vmsin(ωt)V_m \sin(\omega t)Vmsin(ωt). The average load voltage is:
Correct Answer: c. Vm2π⋅(1+cosα)\frac{V_m}{2\pi} \cdot (1 + \cos\alpha)2πVm⋅(1+cosα)
Explanation: The average voltage is calculated using integration over one conduction period and considering the firing angle α\alphaα. This formula is standard for such choppers: Vavg=Vm2π⋅(1+cosα)V_{\text{avg}} = \frac{V_m}{2\pi} \cdot (1 + \cos\alpha)Vavg=2πVm⋅(1+cosα)
Question 15
In a three-phase three-pulse rectifier with resistive load, SCR T1T_1T1 conducts first. If T1T_1T1 is fired at an angle α>30∘\alpha > 30^\circα>30∘, then T1T_1T1 would conduct from:
Correct Answer: c. 30∘+α to 180∘30^\circ + \alpha \, \text{to} \, 180^\circ30∘+αto180∘
Explanation: In a three-phase three-pulse rectifier, each SCR conducts for 150°. When T1T_1T1 is triggered at an angle α\alphaα, conduction starts at 30∘+α30^\circ + \alpha30∘+α and ends at 180∘180^\circ180∘ as the conduction period is fixed.