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The web-text records are organized with fields such as knowledge domain, question type, difficulty, key concepts, and reasoning steps for targeted search and filtering. It supports supervised fine-tuning of power engineering question-answering models, knowledge coverage evaluation, and the development of learning support tools.\u003C\u002Fp>","2026-10-15 18:04:17","2026-10-19 17:18:37","0.00",0,5000,"power system relay protection dataset,relay protection Q&A,electrical engineering question bank,relay protection practice questions,power engineering knowledge QA,fault identification training data,relay protection SFT dataset,secondary circuit questions","Practice Q&A on relay protection principles, fault identification, setting logic, and secondary circuits, with answers and explanations for model tuning and","JSON","0.3GB","Power System Protection, Electrical Engineering, Technical Question Answering","Question Answering, Knowledge Reasoning, Explanation Generation","V1.0","rivo\u002Fsample-packages\u002F137adab2-f07d-4933-8c6b-1f96fa7384a1\u002F20261015.zip",[583,588,593,599,604,609,614,619,624,629,634],{"id":572,"name":584,"field":585,"intro":586,"type":587},"File Name","file_name","File name","string",{"id":589,"name":590,"field":591,"intro":592,"type":587},8950,"Answer Text","answer_text","Original answer to the practice question, which may contain an option, judgment, calculation result, or short-answer conclusion.",{"id":594,"name":595,"field":596,"intro":597,"type":598},8951,"Key Concepts","key_concepts","Key relay protection concepts, equipment, parameters, or criteria extracted from the question, answer, and explanation.","array",{"id":600,"name":601,"field":602,"intro":603,"type":587},8952,"SFT Response Text","sft_response","A complete supervised fine-tuning target response that combines the answer and explanation with professional reasoning.",{"id":605,"name":606,"field":607,"intro":608,"type":587},8953,"Question Text","question_text","Original text of a power system relay protection practice question.",{"id":610,"name":611,"field":612,"intro":613,"type":587},8954,"Question Type","question_type","Question format inferred from how the problem is presented and answered.",{"id":615,"name":616,"field":617,"intro":618,"type":598},8955,"Reasoning Steps","reasoning_steps","Ordered steps that organize the explanation into judgments, calculations, references to principles, or derivations of the conclusion.",{"id":620,"name":621,"field":622,"intro":623,"type":587},8956,"SFT Instruction Text","sft_instruction","A user question or domain-specific instruction based on the original question and prepared for supervised fine-tuning.",{"id":625,"name":626,"field":627,"intro":628,"type":587},8957,"Difficulty Level","difficulty_level","Question difficulty based on the breadth of knowledge covered, reasoning complexity, and calculation or analysis requirements.",{"id":630,"name":631,"field":632,"intro":633,"type":587},8958,"Explanation Text","explanation_text","Original explanation of the answer, relay protection principles, fault identification criteria, setting logic, or operational procedures.",{"id":635,"name":636,"field":637,"intro":638,"type":587},8959,"Knowledge Domain","knowledge_domain","Primary professional knowledge domain inferred from the question and explanation.",[640,644,648,652],{"id":641,"question":642,"answer":643},6367,"What does the dataset contain?","It includes relay protection practice questions, answers, and explanations, with structured information such as knowledge domain, question type, difficulty level, key concepts, and reasoning steps.",{"id":645,"question":646,"answer":647},6368,"Which model tasks is it suitable for?","It can support domain-specific supervised fine-tuning for power system relay protection Q&A, answer generation, and problem-solving reasoning. It can also help build professional knowledge retrieval and learning support tools.",{"id":649,"question":650,"answer":651},6369,"What are the dataset size and quality metrics?","The available description does not specify a sample count or quality evaluation metrics. Before use, review the product page for sample previews, field completeness, and quality notes, and validate a sample against your target task.",{"id":653,"question":654,"answer":655},6370,"How can I obtain the dataset, and what are the licensing terms?","Check the platform listing for available versions, delivery options, and license terms. Permissions for commercial use, redistribution, and model training depend on the applicable purchase or license agreement.",[657,675,689],{"id":658,"relate_table":659,"relate_id":660,"file_size":572,"file_name":661,"path":8,"duration":572,"cate":483,"file_format":662,"audio_rate":8,"audio_channel":522,"audio_bit_depth":522,"quality":8,"annotation":663,"cate_name":664,"answer_text":665,"key_concepts":666,"sft_response":667,"question_text":668,"question_type":669,"reasoning_steps":670,"sft_instruction":671,"difficulty_level":672,"explanation_text":673,"knowledge_domain":674},15087,"union_cubo",1828685704,"样例_3.json","json","{\"answer_text\": \"The CT ratio is 400\u002F5=80. (1) The primary-side pickup current is 1.25 x 80=100 A; (2) the secondary current corresponding to the 600 A primary short-circuit current is 600\u002F80=7.5 A; (3) 7.5 A is greater than 1.25 A, so the short-circuit current reaches the pickup value of the overcurrent protection.\", \"key_concepts\": \"[\\\"Current transformer ratio\\\", \\\"Overcurrent protection\\\", \\\"Primary pickup current\\\", \\\"Secondary current conversion\\\", \\\"Operating condition\\\", \\\"Operating time delay\\\"]\", \"sft_response\": \"The CT ratio is K_I=400\u002F5=80. The primary-side pickup current is I_{set,1}=1.25 x 80=100 A; the 600 A primary short-circuit current referred to the secondary side is I_{f,2}=600\u002F80=7.5 A. Because 7.5 A is greater than the relay secondary pickup setting value 1.25 A, the overcurrent protection meets the pickup condition. Note that whether it ultimately trips still needs further determination in combination with the operating time delay, trip output circuit, and blocking conditions.\", \"question_text\": \"A 10 kV feeder is equipped with overcurrent protection, the current transformer ratio is 400\u002F5, and the secondary current corresponding to the relay primary-side equivalent current setting is 1.25 A. If a three-phase short circuit occurs on the feeder and the short-circuit current referred to the CT primary side is 600 A, calculate: (1) the primary-side pickup current of the protection; (2) the CT secondary current corresponding to the 600 A primary short-circuit current; (3) determine whether the fault current reaches the protection pickup value.\", \"question_type\": \"Calculation Question\", \"reasoning_steps\": \"[\\\"Calculate the CT ratio: K_I=400\u002F5=80.\\\", \\\"Multiply the relay secondary setting value 1.25 A by the ratio to obtain the primary-side pickup current 100 A.\\\", \\\"Divide the primary-side short-circuit current 600 A by the ratio to obtain the secondary-side fault current 7.5 A.\\\", \\\"Compare 7.5 A with the secondary pickup setting value 1.25 A and confirm that the fault current meets the pickup condition.\\\", \\\"Explain that reaching the pickup value is not equivalent to immediate tripping; the operating time delay and trip output logic still need to be checked.\\\"]\", \"sft_instruction\": \"Given that the CT ratio of a 10 kV feeder is 400\u002F5, the secondary setting current of the overcurrent relay is 1.25 A, and the primary short-circuit current during a three-phase short circuit is 600 A. Calculate the primary pickup current and the secondary current during the fault, and determine whether the protection picks up.\", \"difficulty_level\": \"Medium\", \"explanation_text\": \"The current transformer ratio is K_I=I_{1N}\u002FI_{2N}=400\u002F5=80. When the relay secondary setting current is referred to the primary side, it should be multiplied by the CT ratio, so the primary pickup current is I_{set,1}=1.25 x 80=100 A. When the fault current is referred to the secondary side, it should be divided by the CT ratio, giving I_{f,2}=600\u002F80=7.5 A. Since 7.5 A exceeds the relay secondary pickup setting of 1.25 A, the protection element meets the pickup condition; actual tripping still needs to be determined in combination with the operating time delay, trip output logic, and other blocking conditions.\", \"knowledge_domain\": \"Protection Setting and Coordination\"}","文本","The CT ratio is 400\u002F5=80. (1) The primary-side pickup current is 1.25 x 80=100 A; (2) the secondary current corresponding to the 600 A primary short-circuit current is 600\u002F80=7.5 A; (3) 7.5 A is greater than 1.25 A, so the short-circuit current reaches the pickup value of the overcurrent protection.","[\"Current transformer ratio\", \"Overcurrent protection\", \"Primary pickup current\", \"Secondary current conversion\", \"Operating condition\", \"Operating time delay\"]","The CT ratio is K_I=400\u002F5=80. The primary-side pickup current is I_{set,1}=1.25 x 80=100 A; the 600 A primary short-circuit current referred to the secondary side is I_{f,2}=600\u002F80=7.5 A. Because 7.5 A is greater than the relay secondary pickup setting value 1.25 A, the overcurrent protection meets the pickup condition. Note that whether it ultimately trips still needs further determination in combination with the operating time delay, trip output circuit, and blocking conditions.","A 10 kV feeder is equipped with overcurrent protection, the current transformer ratio is 400\u002F5, and the secondary current corresponding to the relay primary-side equivalent current setting is 1.25 A. If a three-phase short circuit occurs on the feeder and the short-circuit current referred to the CT primary side is 600 A, calculate: (1) the primary-side pickup current of the protection; (2) the CT secondary current corresponding to the 600 A primary short-circuit current; (3) determine whether the fault current reaches the protection pickup value.","Calculation Question","[\"Calculate the CT ratio: K_I=400\u002F5=80.\", \"Multiply the relay secondary setting value 1.25 A by the ratio to obtain the primary-side pickup current 100 A.\", \"Divide the primary-side short-circuit current 600 A by the ratio to obtain the secondary-side fault current 7.5 A.\", \"Compare 7.5 A with the secondary pickup setting value 1.25 A and confirm that the fault current meets the pickup condition.\", \"Explain that reaching the pickup value is not equivalent to immediate tripping; the operating time delay and trip output logic still need to be checked.\"]","Given that the CT ratio of a 10 kV feeder is 400\u002F5, the secondary setting current of the overcurrent relay is 1.25 A, and the primary short-circuit current during a three-phase short circuit is 600 A. Calculate the primary pickup current and the secondary current during the fault, and determine whether the protection picks up.","Medium","The current transformer ratio is K_I=I_{1N}\u002FI_{2N}=400\u002F5=80. When the relay secondary setting current is referred to the primary side, it should be multiplied by the CT ratio, so the primary pickup current is I_{set,1}=1.25 x 80=100 A. When the fault current is referred to the secondary side, it should be divided by the CT ratio, giving I_{f,2}=600\u002F80=7.5 A. Since 7.5 A exceeds the relay secondary pickup setting of 1.25 A, the protection element meets the pickup condition; actual tripping still needs to be determined in combination with the operating time delay, trip output logic, and other blocking conditions.","Protection Setting and Coordination",{"id":676,"relate_table":659,"relate_id":677,"file_size":572,"file_name":678,"path":8,"duration":572,"cate":483,"file_format":662,"audio_rate":8,"audio_channel":522,"audio_bit_depth":522,"quality":8,"annotation":679,"cate_name":664,"answer_text":680,"key_concepts":681,"sft_response":682,"question_text":683,"question_type":684,"reasoning_steps":685,"sft_instruction":686,"difficulty_level":672,"explanation_text":687,"knowledge_domain":688},15086,1318363260,"样例_2.json","{\"answer_text\": \"Direct opening is strictly prohibited. When the CT secondary circuit is open, the secondary current is forced to interrupt, the core magnetic flux increases significantly, and a dangerous high voltage may be generated on the secondary side, endangering personal safety and possibly causing CT insulation damage, increased core residual magnetism, and failure of protection and measurement circuits. Before handling, power should be cut off, isolation should be applied, or dedicated short-circuiting measures should be used in accordance with site regulations, and relevant protection and safety measures should be confirmed.\", \"key_concepts\": \"[\\\"Current transformer\\\", \\\"CT secondary circuit open\\\", \\\"Secondary induced voltage\\\", \\\"Core magnetic flux\\\", \\\"Protection circuit failure\\\", \\\"Dedicated short-circuiting measures\\\"]\", \"sft_response\": \"When a CT operates normally, the secondary circuit should remain closed, and the secondary current is used to offset the magnetomotive force generated by the primary current. If the secondary circuit is directly opened, the secondary current is interrupted, the core magnetic flux and secondary induced voltage may rise significantly, dangerous high voltage may appear at the open point, and CT insulation damage, increased residual magnetism, and failure of protection and measurement circuits may result. Therefore, the CT secondary circuit must not be directly disconnected when there is current in the primary. Before handling, power outage, isolation, or qualified dedicated short-circuiting measures should be carried out in accordance with on-site regulations, and the impact on relevant protection and measurement circuits should be verified; ordinary wires must not be used for arbitrary short-circuiting or operation.\", \"question_text\": \"During operation of a substation, abnormal contact was found at the secondary circuit terminals of a group of current transformers (CTs) in operation, and the terminals need to be handled. Question: Why is it strictly prohibited to directly open the CT secondary circuit when the primary equipment is energized and current is flowing in the CT primary? What should be focused on when handling such defects?\", \"question_type\": \"Fault Analysis Question\", \"reasoning_steps\": \"[\\\"Identify the fault object as the secondary circuit of an operating CT, not an ordinary low-voltage control circuit.\\\", \\\"Analyze that when the CT works normally, the secondary circuit is closed and has low impedance, and the secondary current has a balancing effect on the primary magnetomotive force.\\\", \\\"Determine that after the secondary circuit opens, the secondary current is interrupted, and the core magnetic flux and secondary induced voltage may rise significantly.\\\", \\\"From this, conclude that direct opening may cause personal electric shock, equipment insulation damage, and abnormal protection and measurement.\\\", \\\"During handling, power outage, isolation, or reliable short-circuiting should be completed according to regulations, and the impact on protection devices should be evaluated.\\\"]\", \"sft_instruction\": \"Analyze why it is strictly prohibited to directly open the CT secondary circuit when there is current in the CT primary, and explain the basic safety principles that should be followed when handling abnormal CT secondary terminals.\", \"difficulty_level\": \"Medium\", \"explanation_text\": \"During normal operation, the magnetomotive force generated by the CT secondary winding is used to offset the magnetomotive force generated by the primary current, and the secondary circuit is approximately in a low-impedance closed state. If the secondary circuit suddenly opens, the secondary current cannot flow, the core magnetic flux and secondary induced electromotive force may rise significantly, and dangerous voltage may also appear at the open point. Therefore, the secondary terminals of a loaded CT must not be directly disconnected or opened. Actual handling must comply with on-site operating regulations, and qualified personnel must confirm the primary system status, the scope of protection impact, and short-circuiting and isolation measures; short-circuiting should use qualified test terminals or dedicated short-circuiting tools, and ordinary wires must not be used for arbitrary operation.\", \"knowledge_domain\": \"Secondary Circuit\"}","Direct opening is strictly prohibited. When the CT secondary circuit is open, the secondary current is forced to interrupt, the core magnetic flux increases significantly, and a dangerous high voltage may be generated on the secondary side, endangering personal safety and possibly causing CT insulation damage, increased core residual magnetism, and failure of protection and measurement circuits. Before handling, power should be cut off, isolation should be applied, or dedicated short-circuiting measures should be used in accordance with site regulations, and relevant protection and safety measures should be confirmed.","[\"Current transformer\", \"CT secondary circuit open\", \"Secondary induced voltage\", \"Core magnetic flux\", \"Protection circuit failure\", \"Dedicated short-circuiting measures\"]","When a CT operates normally, the secondary circuit should remain closed, and the secondary current is used to offset the magnetomotive force generated by the primary current. If the secondary circuit is directly opened, the secondary current is interrupted, the core magnetic flux and secondary induced voltage may rise significantly, dangerous high voltage may appear at the open point, and CT insulation damage, increased residual magnetism, and failure of protection and measurement circuits may result. Therefore, the CT secondary circuit must not be directly disconnected when there is current in the primary. Before handling, power outage, isolation, or qualified dedicated short-circuiting measures should be carried out in accordance with on-site regulations, and the impact on relevant protection and measurement circuits should be verified; ordinary wires must not be used for arbitrary short-circuiting or operation.","During operation of a substation, abnormal contact was found at the secondary circuit terminals of a group of current transformers (CTs) in operation, and the terminals need to be handled. Question: Why is it strictly prohibited to directly open the CT secondary circuit when the primary equipment is energized and current is flowing in the CT primary? What should be focused on when handling such defects?","Fault Analysis Question","[\"Identify the fault object as the secondary circuit of an operating CT, not an ordinary low-voltage control circuit.\", \"Analyze that when the CT works normally, the secondary circuit is closed and has low impedance, and the secondary current has a balancing effect on the primary magnetomotive force.\", \"Determine that after the secondary circuit opens, the secondary current is interrupted, and the core magnetic flux and secondary induced voltage may rise significantly.\", \"From this, conclude that direct opening may cause personal electric shock, equipment insulation damage, and abnormal protection and measurement.\", \"During handling, power outage, isolation, or reliable short-circuiting should be completed according to regulations, and the impact on protection devices should be evaluated.\"]","Analyze why it is strictly prohibited to directly open the CT secondary circuit when there is current in the CT primary, and explain the basic safety principles that should be followed when handling abnormal CT secondary terminals.","During normal operation, the magnetomotive force generated by the CT secondary winding is used to offset the magnetomotive force generated by the primary current, and the secondary circuit is approximately in a low-impedance closed state. If the secondary circuit suddenly opens, the secondary current cannot flow, the core magnetic flux and secondary induced electromotive force may rise significantly, and dangerous voltage may also appear at the open point. Therefore, the secondary terminals of a loaded CT must not be directly disconnected or opened. Actual handling must comply with on-site operating regulations, and qualified personnel must confirm the primary system status, the scope of protection impact, and short-circuiting and isolation measures; short-circuiting should use qualified test terminals or dedicated short-circuiting tools, and ordinary wires must not be used for arbitrary operation.","Secondary Circuit",{"id":690,"relate_table":659,"relate_id":691,"file_size":572,"file_name":692,"path":8,"duration":572,"cate":483,"file_format":662,"audio_rate":8,"audio_channel":522,"audio_bit_depth":522,"quality":8,"annotation":693,"cate_name":664,"answer_text":694,"key_concepts":695,"sft_response":696,"question_text":697,"question_type":698,"reasoning_steps":699,"sft_instruction":700,"difficulty_level":701,"explanation_text":702,"knowledge_domain":703},15085,1914734940,"样例_1.json","{\"answer_text\": \"B。\", \"key_concepts\": \"[\\\"Distance protection\\\", \\\"Measured impedance\\\", \\\"Distance Zone I\\\", \\\"Protection range\\\", \\\"Backup protection\\\"]\", \"sft_response\": \"The correct option is B. Distance protection determines the fault distance based on measured impedance. To ensure selectivity, distance Zone I is usually set to 80%-85% of the total length of the protected line and usually has no intentional time delay; distance Zone II extends to a part of the adjacent line and is set with a certain time delay; distance Zone III serves as backup protection for the protected line and adjacent elements, and its operating time delay is usually longer.\", \"question_text\": \"Regarding three-zone distance protection for a 110 kV transmission line, which of the following statements is correct: A. Distance Zone I usually protects the entire line length and has a long time delay; B. Distance Zone I is usually set to 80%-85% of the total line length and usually operates without intentional time delay; C. Distance Zone II protects only the local end of the protected line and must operate without time delay; D. Distance Zone III is used only to protect the busbar.\", \"question_type\": \"Single-choice question\", \"reasoning_steps\": \"[\\\"Clarify that distance protection reflects the fault distance through measured impedance.\\\", \\\"Considering measurement errors, fault resistance, and external fault effects, Zone I cannot cover the entire line length without sacrificing selectivity.\\\", \\\"According to common setting principles, the protection range of Zone I is usually 80%-85% of the total line length.\\\", \\\"After checking each option, option B conforms to the zone-setting principle of distance protection.\\\"]\", \"sft_instruction\": \"Determine which statement about three-zone distance protection for a 110 kV transmission line is correct, and explain the basic characteristics of distance Zones I, II, and III in terms of protection range and operating time delay.\", \"difficulty_level\": \"Basic\", \"explanation_text\": \"Distance protection determines the fault location based on the measured fault impedance. To avoid the influence of factors such as external faults at the remote busbar, measurement errors, and fault resistance, distance Zone I usually protects only about 80%-85% of the total length of the protected line and operates with a short inherent time or approximately without time delay. Distance Zone II should extend to a part of the adjacent line and have a time delay, while distance Zone III mainly serves as backup protection for the protected line and adjacent elements.\", \"knowledge_domain\": \"Relay Protection Principles\"}","B。","[\"Distance protection\", \"Measured impedance\", \"Distance Zone I\", \"Protection range\", \"Backup protection\"]","The correct option is B. Distance protection determines the fault distance based on measured impedance. To ensure selectivity, distance Zone I is usually set to 80%-85% of the total length of the protected line and usually has no intentional time delay; distance Zone II extends to a part of the adjacent line and is set with a certain time delay; distance Zone III serves as backup protection for the protected line and adjacent elements, and its operating time delay is usually longer.","Regarding three-zone distance protection for a 110 kV transmission line, which of the following statements is correct: A. Distance Zone I usually protects the entire line length and has a long time delay; B. Distance Zone I is usually set to 80%-85% of the total line length and usually operates without intentional time delay; C. Distance Zone II protects only the local end of the protected line and must operate without time delay; D. Distance Zone III is used only to protect the busbar.","Single-choice question","[\"Clarify that distance protection reflects the fault distance through measured impedance.\", \"Considering measurement errors, fault resistance, and external fault effects, Zone I cannot cover the entire line length without sacrificing selectivity.\", \"According to common setting principles, the protection range of Zone I is usually 80%-85% of the total line length.\", \"After checking each option, option B conforms to the zone-setting principle of distance protection.\"]","Determine which statement about three-zone distance protection for a 110 kV transmission line is correct, and explain the basic characteristics of distance Zones I, II, and III in terms of protection range and operating time delay.","Basic","Distance protection determines the fault location based on the measured fault impedance. To avoid the influence of factors such as external faults at the remote busbar, measurement errors, and fault resistance, distance Zone I usually protects only about 80%-85% of the total length of the protected line and operates with a short inherent time or approximately without time delay. Distance Zone II should extend to a part of the adjacent line and have a time delay, while distance Zone III mainly serves as backup protection for the protected line and adjacent elements.","Relay Protection Principles","{\"answer_text\":\"The CT ratio is 400\\\u002F5=80. (1) The primary-side pickup current is 1.25 x 80=100 A; (2) the secondary current corresponding to the 600 A primary short-circuit current is 600\\\u002F80=7.5 A; (3) 7.5 A is greater than 1.25 A, so the short-circuit current reaches the pickup value of the overcurrent protection.\",\"key_concepts\":\"[\\\"Current transformer ratio\\\", \\\"Overcurrent protection\\\", \\\"Primary pickup current\\\", \\\"Secondary current conversion\\\", \\\"Operating condition\\\", \\\"Operating time delay\\\"]\",\"sft_response\":\"The CT ratio is K_I=400\\\u002F5=80. The primary-side pickup current is I_{set,1}=1.25 x 80=100 A; the 600 A primary short-circuit current referred to the secondary side is I_{f,2}=600\\\u002F80=7.5 A. Because 7.5 A is greater than the relay secondary pickup setting value 1.25 A, the overcurrent protection meets the pickup condition. Note that whether it ultimately trips still needs further determination in combination with the operating time delay, trip output circuit, and blocking conditions.\",\"question_text\":\"A 10 kV feeder is equipped with overcurrent protection, the current transformer ratio is 400\\\u002F5, and the secondary current corresponding to the relay primary-side equivalent current setting is 1.25 A. If a three-phase short circuit occurs on the feeder and the short-circuit current referred to the CT primary side is 600 A, calculate: (1) the primary-side pickup current of the protection; (2) the CT secondary current corresponding to the 600 A primary short-circuit current; (3) determine whether the fault current reaches the protection pickup value.\",\"question_type\":\"Calculation Question\",\"reasoning_steps\":\"[\\\"Calculate the CT ratio: K_I=400\\\u002F5=80.\\\", \\\"Multiply the relay secondary setting value 1.25 A by the ratio to obtain the primary-side pickup current 100 A.\\\", \\\"Divide the primary-side short-circuit current 600 A by the ratio to obtain the secondary-side fault current 7.5 A.\\\", \\\"Compare 7.5 A with the secondary pickup setting value 1.25 A and confirm that the fault current meets the pickup condition.\\\", \\\"Explain that reaching the pickup value is not equivalent to immediate tripping; the operating time delay and trip output logic still need to be checked.\\\"]\",\"sft_instruction\":\"Given that the CT ratio of a 10 kV feeder is 400\\\u002F5, the secondary setting current of the overcurrent relay is 1.25 A, and the primary short-circuit current during a three-phase short circuit is 600 A. Calculate the primary pickup current and the secondary current during the fault, and determine whether the protection picks up.\",\"difficulty_level\":\"Medium\",\"explanation_text\":\"The current transformer ratio is K_I=I_{1N}\\\u002FI_{2N}=400\\\u002F5=80. When the relay secondary setting current is referred to the primary side, it should be multiplied by the CT ratio, so the primary pickup current is I_{set,1}=1.25 x 80=100 A. When the fault current is referred to the secondary side, it should be divided by the CT ratio, giving I_{f,2}=600\\\u002F80=7.5 A. Since 7.5 A exceeds the relay secondary pickup setting of 1.25 A, the protection element meets the pickup condition; actual tripping still needs to be determined in combination with the operating time delay, trip output logic, and other blocking conditions.\",\"knowledge_domain\":\"Protection Setting and Coordination\"}","MOBIUSI INC","https:\u002F\u002Fstatic.mobiusi.com\u002Fmobiwusi\u002F20250829\u002F7bf4ac62b4533e2b809998a6f1c5491e.png?Expires=1792463130&OSSAccessKeyId=LTAI5tH8MQqhgeiAtrzCHrxM&Signature=v6Tr8jBrdm4s5wU1SnnyGsYg%2Bhw%3D&x-oss-process=image%2Fresize%2Cm_fill%2Cw_35%2Ch_35",[708,714],{"type":709,"title":710,"count_data":711},"bar","Length distribution",[712],{"name":713,"value":483},"1k lower",{"type":715,"title":716,"count_data":717},"pie","Language distribution",[718],{"name":719,"value":483},"English"]