Single Phase Half Wave Converter Drives MCQ Quiz - Objective Question with Answer for Single Phase Half Wave Converter Drives - Download Free PDF
Last updated on Jun 23, 2025
Latest Single Phase Half Wave Converter Drives MCQ Objective Questions
Single Phase Half Wave Converter Drives Question 1:
The four quadrant separately excited by a DC motor shown in the figure is powered from a 100 V battery. The field current is kept constant. The armature resistance is 0.6 Ω. When the converter is operating at a duty cycle of 0.8 and armature is drawing constant current of 30 A with negligible ripple, what is the back emf developed across the coil?
Answer (Detailed Solution Below)
Single Phase Half Wave Converter Drives Question 1 Detailed Solution
Explanation:
Step-by-Step Solution:
- Understanding the Duty Cycle:
- The duty cycle is the fraction of time the converter conducts during one period. A duty cycle of 0.8 implies that the converter is ON for 80% of the time and OFF for the remaining 20%.
- The equivalent voltage applied to the armature can be calculated using the duty cycle and battery voltage.
- Voltage Applied to the Armature:
- The converter modifies the battery voltage depending on the duty cycle. The applied voltage (Vapplied) is given by:
- Vapplied = Duty Cycle × Battery Voltage
- Substituting the given values:
- Vapplied = 0.8 × 100 V = 80 V
- Armature Voltage Equation:
- The voltage applied to the armature is distributed across the armature resistance and the back emf (Eb). The armature voltage equation is:
- Vapplied = Eb + Ia × Ra
- Where:
- Eb = Back emf developed across the motor
- Ia = Armature current (30 A)
- Ra = Armature resistance (0.6 Ω)
- Substitute the Known Values:
- From the equation:
- 80 V = Eb + (30 A × 0.6 Ω)
- Simplifying the resistance term:
- 30 A × 0.6 Ω = 18 V
- Therefore:
- 80 V = Eb + 18 V
- Solve for Eb:
- Eb = 80 V - 18 V = 62 V
Back emf developed across the coil:
The back emf across the coil is calculated to be 62 V
Top Single Phase Half Wave Converter Drives MCQ Objective Questions
The four quadrant separately excited by a DC motor shown in the figure is powered from a 100 V battery. The field current is kept constant. The armature resistance is 0.6 Ω. When the converter is operating at a duty cycle of 0.8 and armature is drawing constant current of 30 A with negligible ripple, what is the back emf developed across the coil?
Answer (Detailed Solution Below)
Single Phase Half Wave Converter Drives Question 2 Detailed Solution
Download Solution PDFExplanation:
Step-by-Step Solution:
- Understanding the Duty Cycle:
- The duty cycle is the fraction of time the converter conducts during one period. A duty cycle of 0.8 implies that the converter is ON for 80% of the time and OFF for the remaining 20%.
- The equivalent voltage applied to the armature can be calculated using the duty cycle and battery voltage.
- Voltage Applied to the Armature:
- The converter modifies the battery voltage depending on the duty cycle. The applied voltage (Vapplied) is given by:
- Vapplied = Duty Cycle × Battery Voltage
- Substituting the given values:
- Vapplied = 0.8 × 100 V = 80 V
- Armature Voltage Equation:
- The voltage applied to the armature is distributed across the armature resistance and the back emf (Eb). The armature voltage equation is:
- Vapplied = Eb + Ia × Ra
- Where:
- Eb = Back emf developed across the motor
- Ia = Armature current (30 A)
- Ra = Armature resistance (0.6 Ω)
- Substitute the Known Values:
- From the equation:
- 80 V = Eb + (30 A × 0.6 Ω)
- Simplifying the resistance term:
- 30 A × 0.6 Ω = 18 V
- Therefore:
- 80 V = Eb + 18 V
- Solve for Eb:
- Eb = 80 V - 18 V = 62 V
Back emf developed across the coil:
The back emf across the coil is calculated to be 62 V
Single Phase Half Wave Converter Drives Question 3:
A separately excited dc motor shown in the figure has back emf constant of 0.20 V/rpm. The armature current is 6 A without any ripple and armature resistance of 3 Ω. Find the speed of the motor (in RPM) at a firing angle of 30°.
Answer (Detailed Solution Below) 410 - 420
Single Phase Half Wave Converter Drives Question 3 Detailed Solution
Average output voltage of 1ϕ half controlled rectifier = Vo
\({V_o} = \frac{{{V_m}}}{{2\pi }}\left( {1 + cos \alpha } \right) = \frac{{240\sqrt 2 }}{{2\pi }}\left( {1 + cos30^\circ } \right)\)
Vo = 100.85 V
Ea = 100.85 – 6 × 3 = 82.85 V
Ea = kbN
\(N = \frac{{82.85}}{{0.20}} = 414.25\;RPM\)
Single Phase Half Wave Converter Drives Question 4:
The four quadrant separately excited by a DC motor shown in the figure is powered from a 100 V battery. The field current is kept constant. The armature resistance is 0.6 Ω. When the converter is operating at a duty cycle of 0.8 and armature is drawing constant current of 30 A with negligible ripple, what is the back emf developed across the coil?
Answer (Detailed Solution Below)
Single Phase Half Wave Converter Drives Question 4 Detailed Solution
Explanation:
Step-by-Step Solution:
- Understanding the Duty Cycle:
- The duty cycle is the fraction of time the converter conducts during one period. A duty cycle of 0.8 implies that the converter is ON for 80% of the time and OFF for the remaining 20%.
- The equivalent voltage applied to the armature can be calculated using the duty cycle and battery voltage.
- Voltage Applied to the Armature:
- The converter modifies the battery voltage depending on the duty cycle. The applied voltage (Vapplied) is given by:
- Vapplied = Duty Cycle × Battery Voltage
- Substituting the given values:
- Vapplied = 0.8 × 100 V = 80 V
- Armature Voltage Equation:
- The voltage applied to the armature is distributed across the armature resistance and the back emf (Eb). The armature voltage equation is:
- Vapplied = Eb + Ia × Ra
- Where:
- Eb = Back emf developed across the motor
- Ia = Armature current (30 A)
- Ra = Armature resistance (0.6 Ω)
- Substitute the Known Values:
- From the equation:
- 80 V = Eb + (30 A × 0.6 Ω)
- Simplifying the resistance term:
- 30 A × 0.6 Ω = 18 V
- Therefore:
- 80 V = Eb + 18 V
- Solve for Eb:
- Eb = 80 V - 18 V = 62 V
Back emf developed across the coil:
The back emf across the coil is calculated to be 62 V
Single Phase Half Wave Converter Drives Question 5:
Consider a 1ϕ half wave uncontrolled rectifier with the pure inductive load as shown in the figure below:
The maximum value of load current is:
Answer (Detailed Solution Below)
Single Phase Half Wave Converter Drives Question 5 Detailed Solution
Calculation:
During +ve half cycle from 0 < Vin < π, SCR conducts.
\(V_o= V_s\)
\(L {di_o(t) \over dt}=V_m\space sin\space ω t\)
\(i_o(t)={1\over L}\int_{0}^{t}V_m\space sin\space ω t\space dt\)
\(i_o(t)={V_m\over ω L}(-cos\space ω t)_{0}^{t}\)
\(i_o(t)={V_m\over ω L}(1-cos\space ω t)\)
The maximum value of load current occurs at ωt =π
\(i_o(\pi)={V_m\over ω L}(1-cos\space \pi)\)
\(i_o(t)_{max} = {2V_m\over \omega L}\)
Single Phase Half Wave Converter Drives Question 6:
A separately excited dc motor is supplied from 230 V, 50 Hz source through a single-phase half-wave-controlled converter. It field is fed through 1-phase semi converter with zero-degree firing angle delay. Motor resistance ra = 0.8 Ω and motor constant = 0.6 V-sec/rad. For and motor constant = 0.6 V-sec/rad. For rated load torque of 20 N-m at 1000 rpm and for continuous ripple free current, input power factor.
Answer (Detailed Solution Below) 0.85 - 0.95
Single Phase Half Wave Converter Drives Question 6 Detailed Solution
Armature current \(= \frac{{Te}}{{{K_m}}} = \frac{{20}}{{0.6}} = 33.33\;A\)
Motor EMF Ea \(= {K_m}.{\omega _m} = 0.6 \times \frac{{2\pi \times 1000}}{{60}}\) = 62.8 V
\({V_t} = \frac{{{V_m}}}{{2\pi }}\left( {1 + \cos \propto } \right) = {E_a} + {I_a}{r_a}\)
\(\frac{{\sqrt 2 \times 230}}{{2\pi }}\left( {1 + \cos \propto } \right) = 62.8 + 33.33 \times 0.8\)
51.79 (1 + cos ∝) = 89.464
1 + cos ∝ = 1.7272
∝ = 43.35°
Input power factor \( = \frac{{\sqrt 2 \left( {1 + \cos \alpha } \right)}}{{\sqrt {\pi \left( {\pi - \alpha } \right)} }} = \frac{{\sqrt 2 \times 1.7272}}{{2.718}}\)
= 0.898 lag