Question
Download Solution PDFA 16 kV DC source having an internal resistance of 1 ohm supplies 900A to a 12 kV, 3 phase 6 pulse 60 Hz inverter. Calculate the DC voltage generated by the inverter.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Calculation of DC Voltage Generated by the Inverter
To calculate the DC voltage generated by the inverter, let us analyze the given data step-by-step:
- DC source voltage, Vsource = 16 kV = 16000 V
- Internal resistance of the DC source, Rinternal = 1 Ω
- DC current supplied by the source, IDC = 900 A
- Inverter output voltage, Vinverter = 12 kV = 12000 V
Step 1: Voltage Drop Across the Internal Resistance
The internal resistance of the DC source causes a voltage drop, which can be calculated using Ohm's Law:
Voltage Drop (Vdrop) = IDC × Rinternal
Substitute the given values:
Vdrop = 900 A × 1 Ω = 900 V
Step 2: DC Voltage at the Terminals of the Source
The DC voltage at the terminals of the source is the source voltage minus the voltage drop across the internal resistance:
Vterminals = Vsource - Vdrop
Substitute the values:
Vterminals = 16000 V - 900 V = 15100 V
Step 3: DC Voltage Generated by the Inverter
The inverter generates a DC voltage that corresponds to the terminal voltage of the source because it is directly connected to the source.
Thus, the DC voltage generated by the inverter is:
Vinverter = Vterminals = 15100 V
Final Answer: The DC voltage generated by the inverter is 15100 V.
Analysis of Other Options:
Let us now analyze why the other options are incorrect:
Option 1: 27000 V
This option is incorrect because the DC source voltage is only 16000 V, and the internal resistance causes a voltage drop. The DC voltage generated by the inverter cannot exceed the source voltage. The value of 27000 V is unrealistic and does not match the calculations.
Option 2: 15100 V
This is the correct answer as derived above. The terminal voltage of the DC source, after accounting for the voltage drop across the internal resistance, is 15100 V, which is the voltage generated by the inverter.
Option 3: 16000 V
This option assumes that there is no voltage drop across the internal resistance of the DC source. However, the problem explicitly states that the internal resistance is 1 Ω and the current is 900 A, leading to a voltage drop of 900 V. Thus, the terminal voltage is less than 16000 V, making this option incorrect.
Option 4: 16400 V
This option is incorrect as it exceeds the source voltage of 16000 V. Additionally, it does not account for the voltage drop caused by the internal resistance of the source. Such a value is not physically possible in this scenario.
Conclusion:
The correct answer is Option 2: 15100 V. This value accurately accounts for the voltage drop across the internal resistance of the DC source and represents the DC voltage generated by the inverter.
Additional Information
Important Considerations:
- The internal resistance of the DC source plays a significant role in determining the terminal voltage. For high currents, even small resistances can cause substantial voltage drops.
- In practical applications, minimizing the internal resistance of the source is crucial to maximize the efficiency of the system.
- Six-pulse inverters are commonly used in industrial applications due to their simplicity and cost-effectiveness. However, they may introduce harmonics into the system, which need to be mitigated using filters.
Last updated on Jul 1, 2025
-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been rescheduled on its official website.
-> JKSSB JE exam will be conducted on 31st August (Civil), and on 24th August 2025 (Electrical).
-> JKSSB JE application form correction facility has been started. Candidates can make corrections in the JKSSB recruitment 2025 form from June 23 to 27.
-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering.
-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025.
-> JKSSB JE Online Application form will be activated from 18th May 2025 to 16th June 2025
-> Candidates who are preparing for the exam can access the JKSSB JE syllabus PDF from official website of JKSSB.
-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.
-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.