// Numbas version: exam_results_page_options {"name": "PracticeExamForAaronScott", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": true, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "RLC networks", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", ""], "variable_overrides": [[], [], []], "questions": [{"name": "2021_Midterm_6_RLCseries_VI_PQ", "extensions": ["geogebra", "quantities", "weh"], "custom_part_types": [{"source": {"pk": 19, "author": {"name": "William Haynes", "pk": 2530}, "edit_page": "/part_type/19/edit"}, "name": "Engineering Accuracy with units", "short_name": "engineering-answer", "description": "

A value with units marked right if within an adjustable % error of the correct value. Marked close if within a wider margin of error.

", "help_url": "", "input_widget": "string", "input_options": {"correctAnswer": "siground(settings['correctAnswer'],4)", "hint": {"static": true, "value": ""}, "allowEmpty": {"static": true, "value": true}}, "can_be_gap": true, "can_be_step": true, "marking_script": "mark:\nswitch( \n right and good_units and right_sign, add_credit(1.0,'Correct.'),\n right and good_units and not right_sign, add_credit(settings['C2'],'Wrong sign.'),\n right and right_sign and not good_units, add_credit(settings['C2'],'Correct value, but wrong or missing units.'),\n close and good_units, add_credit(settings['C1'],'Close.'),\n close and not good_units, add_credit(settings['C3'],'Answer is close, but wrong or missing units.'),\n incorrect('Wrong answer.')\n)\n\n\ninterpreted_answer:\nqty(student_scalar, student_units)\n\n\n\ncorrect_quantity:\nsettings[\"correctAnswer\"]\n\n\n\ncorrect_units:\nunits(correct_quantity)\n\n\nallowed_notation_styles:\n[\"plain\",\"en\"]\n\nmatch_student_number:\nmatchnumber(studentAnswer,allowed_notation_styles)\n\nstudent_scalar:\nmatch_student_number[1]\n\nstudent_units:\nreplace_regex('ohms','ohm',\n replace_regex('\u00b0', ' deg',\n replace_regex('-', ' ' ,\n studentAnswer[len(match_student_number[0])..len(studentAnswer)])),\"i\")\n\ngood_units:\ntry(\ncompatible(quantity(1, student_units),correct_units),\nmsg,\nfeedback(msg);false)\n\n\nstudent_quantity:\nswitch(not good_units, \n student_scalar * correct_units, \n not right_sign,\n -quantity(student_scalar, student_units),\n quantity(student_scalar,student_units)\n)\n \n\n\npercent_error:\ntry(\nscalar(abs((correct_quantity - student_quantity)/correct_quantity))*100 \n,msg,\nif(student_quantity=correct_quantity,0,100))\n \n\nright:\npercent_error <= settings['right']\n\n\nclose:\nright_sign and percent_error <= settings['close']\n\nright_sign:\nsign(student_scalar) = sign(correct_quantity)", "marking_notes": [{"name": "mark", "description": "This is the main marking note. It should award credit and provide feedback based on the student's answer.", "definition": "switch( \n right and good_units and right_sign, add_credit(1.0,'Correct.'),\n right and good_units and not right_sign, add_credit(settings['C2'],'Wrong sign.'),\n right and right_sign and not good_units, add_credit(settings['C2'],'Correct value, but wrong or missing units.'),\n close and good_units, add_credit(settings['C1'],'Close.'),\n close and not good_units, add_credit(settings['C3'],'Answer is close, but wrong or missing units.'),\n incorrect('Wrong answer.')\n)\n"}, {"name": "interpreted_answer", "description": "A value representing the student's answer to this part.", "definition": "qty(student_scalar, student_units)\n\n"}, {"name": "correct_quantity", "description": "", "definition": "settings[\"correctAnswer\"]\n\n"}, {"name": "correct_units", "description": "", "definition": "units(correct_quantity)\n"}, {"name": "allowed_notation_styles", "description": "", "definition": "[\"plain\",\"en\"]"}, {"name": "match_student_number", "description": "", "definition": "matchnumber(studentAnswer,allowed_notation_styles)"}, {"name": "student_scalar", "description": "", "definition": "match_student_number[1]"}, {"name": "student_units", "description": "

Modify the unit portion of the student's answer by

1. replacing \"ohms\" with \"ohm\" case insensitive

2. replacing '-' with ' '

3. replacing '°' with ' deg'

to allow answers like 10 ft-lb and 30°

", "definition": "replace_regex('ohms','ohm',\n replace_regex('\u00b0', ' deg',\n replace_regex('-', ' ' ,\n studentAnswer[len(match_student_number[0])..len(studentAnswer)])),\"i\")"}, {"name": "good_units", "description": "", "definition": "try(\ncompatible(quantity(1, student_units),correct_units),\nmsg,\nfeedback(msg);false)\n"}, {"name": "student_quantity", "description": "

This fixes the student answer for two common errors.

If student_units are wrong - replace with correct units

If student_scalar has the wrong sign - replace with right sign

If student makes both errors, only one gets fixed.

", "definition": "switch(not good_units, \n student_scalar * correct_units, \n not right_sign,\n -quantity(student_scalar, student_units),\n quantity(student_scalar,student_units)\n)\n \n"}, {"name": "percent_error", "description": "", "definition": "try(\nscalar(abs((correct_quantity - student_quantity)/correct_quantity))*100 \n,msg,\nif(student_quantity=correct_quantity,0,100))\n "}, {"name": "right", "description": "", "definition": "percent_error <= settings['right']\n"}, {"name": "close", "description": "

Only marked close if the student actually has the right sign.

", "definition": "right_sign and percent_error <= settings['close']"}, {"name": "right_sign", "description": "", "definition": "sign(student_scalar) = sign(correct_quantity) "}], "settings": [{"name": "correctAnswer", "label": "Correct Quantity.", "help_url": "", "hint": "The correct answer given as a JME quantity.", "input_type": "code", "default_value": "", "evaluate": true}, {"name": "right", "label": "% Accuracy for right.", "help_url": "", "hint": "Question will be considered correct if the scalar part of the student's answer is within this % of correct value.", "input_type": "code", "default_value": "0.2", "evaluate": true}, {"name": "close", "label": "% Accuracy for close.", "help_url": "", "hint": "Question will be considered close if the scalar part of the student's answer is within this % of correct value.", "input_type": "code", "default_value": "1.0", "evaluate": true}, {"name": "C1", "label": "Close with units.", "help_url": "", "hint": "Partial Credit for close value with appropriate units. if correct answer is 100 N and close is ±1%,
99 N is accepted.", "input_type": "percent", "default_value": "75"}, {"name": "C2", "label": "No units or wrong sign", "help_url": "", "hint": "Partial credit for forgetting units or using wrong sign.
If the correct answer is 100 N, both 100 and -100 N are accepted.", "input_type": "percent", "default_value": "50"}, {"name": "C3", "label": "Close, no units.", "help_url": "", "hint": "Partial Credit for close value but forgotten units.
This value would be close if the expected units were provided. If the correct answer is 100 N, and close is ±1%,
99 is accepted.", "input_type": "percent", "default_value": "25"}], "public_availability": "always", "published": true, "extensions": ["quantities"]}], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "William Haynes", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2530/"}, {"name": "Judy Rea", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/8753/"}], "tags": [], "metadata": {"description": "

series of R and L and C

find Z (mag and Phase)

find I (given V)

Find S, P and Q

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

A circuit consists of a series of resistors,inductors and capacitors is supplied by a {Vs} V ac voltage source.

The reistance is $R= $ {qty({R1},'ohm')}.

The reactances of the inductor and the capacitor are $X_L = $ {qty(XL,'ohm')} and $X_C = ${qty(Xc,'ohm')} .

", "advice": "

1. Find the total Impedance of the network

$Z_T=R+j(X_L-X_C)$

$Z_{Total} = R +j (X_L - X_C)$

$Z_{Total} = \\var{R1} +j (\\var{XL} - \\var{XC}) = \\var{Zt} $

or expressed in polar form

$Z_{Total} = \\var{ZtMag} \\angle \\var{ZtPh} ^{\\circ}$

(Note treat i and j as identical here ... the maths programme can't be edited to use j so we are stuck with i for now, J.Rea 11th Oct 2021)

2. Use Ohms Law

Set 0 deg to be the angle of the voltage. This is our reference phasor. All other phasors will be relative to this one.

So $V_s=\\var{Vs}+0j = \\var{Vs} \\angle \\var{0}^{\\circ}$

$I_s=\\frac{V_s}{Z_t}$ , taking care to include use the full complex number for each $I_s$,$V_s$ and $Z_t$

$I_s=\\frac{\\var{Vs} \\angle \\var{0}^{\\circ}}{\\var{ZtMag} \\angle \\var{ZtPh}^{\\circ}} = \\var{IsMag} \\angle {\\var{IsPh}^{\\circ}}$

3. Use the Power law

$S= V_s I_s^{*}$

$ S = (\\var{Vs} \\angle \\var{0})(\\var{IsMag} \\angle \\var{IsPh}^{\\circ})^{*}$

taking the complex conjugate of the current ... we can find the Total power (also called apparent power).

$ Ss = (\\var{Vs} \\angle \\var{0}^{\\circ})(\\var{IsMag} \\angle \\var{IsPhTimesMinusOne}^{\\circ})$

$ Ss = \\var{SsMag} \\angle \\var{SsPh} MVA = \\var{Ps} + j \\var{Qs} MVA$

Remebering that $S = P +jQ$ we can equate the real and imaginary parts to say

Real power (also known as Active power) is $P=\\var{Ps} MW$

Imaginary power (also known as Reactive power) is $Q=\\var{Qs} MVA_r$

4. Power Factor

Recall the definition of power factor :

$pf = \\frac{P}{|S|} = cos{\\theta}$ where ${\\theta}$ is the angle between the voltage and the current.

$pf \\frac{P}{|S|} =\\frac{\\var{Ps}}{\\var{SsMag}} = \\var{pf}

or using a differnt equation

The phase angle between voltage and current, $\\theta = 0-\\var{IsPh} = \\var{IsPhTimesMinusOne}$

So $pf= cos{\\var{IsPhTimesMinusOne}} = \\var{pf}$

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"Zt": {"name": "Zt", "group": "Ungrouped variables", "definition": "Rt+i*Xt", "description": "", "templateType": "anything", "can_override": false}, "Xl1": {"name": "Xl1", "group": "Ungrouped variables", "definition": "XL", "description": "", "templateType": "anything", "can_override": false}, "Xt": {"name": "Xt", "group": "Ungrouped variables", "definition": "XL1-XC1", "description": "", "templateType": "anything", "can_override": false}, "XC1": {"name": "XC1", "group": "Ungrouped variables", "definition": "XC", "description": "", "templateType": "anything", "can_override": false}, "Rt": {"name": "Rt", "group": "Ungrouped variables", "definition": "R1", "description": "", "templateType": "anything", "can_override": false}, "Vs": {"name": "Vs", "group": "Given in question", "definition": "random(180 .. 280#10)", "description": "", "templateType": "randrange", "can_override": false}, "Is": {"name": "Is", "group": "Ungrouped variables", "definition": "Vs/Zt", "description": "", "templateType": "anything", "can_override": false}, "Ss": {"name": "Ss", "group": "Ungrouped variables", "definition": "Vs*conj(Is)", "description": "", "templateType": "anything", "can_override": false}, "Ps": {"name": "Ps", "group": "Ungrouped variables", "definition": "re(Ss)", "description": "", "templateType": "anything", "can_override": false}, "Qs": {"name": "Qs", "group": "Ungrouped variables", "definition": "im(Ss)", "description": "", "templateType": "anything", "can_override": false}, "pf": {"name": "pf", "group": "Ungrouped variables", "definition": "Ps/abs(Ss)", "description": "", "templateType": "anything", "can_override": false}, "XL": {"name": "XL", "group": "Given in question", "definition": "random(5..100#10 except 0)", "description": "", "templateType": "anything", "can_override": false}, "XC": {"name": "XC", "group": "Given in question", "definition": "random(5..100#10 except 0)", "description": "", "templateType": "anything", "can_override": false}, "R1": {"name": "R1", "group": "Given in question", "definition": "random(50 .. 60#5)", "description": "", "templateType": "randrange", "can_override": false}, "minTol": {"name": "minTol", "group": "FreqUsedDefinitions", "definition": "0.99", "description": "", "templateType": "anything", "can_override": false}, "maxTol": {"name": "maxTol", "group": "FreqUsedDefinitions", "definition": "1.01", "description": "", "templateType": "anything", "can_override": false}, "ZtMag": {"name": "ZtMag", "group": "Variables for use in worked solution", "definition": "abs(Zt)", "description": "", "templateType": "anything", "can_override": false}, "ZtPh": {"name": "ZtPh", "group": "Variables for use in worked solution", "definition": "arg(Zt)*180/pi", "description": "", "templateType": "anything", "can_override": false}, "IsMag": {"name": "IsMag", "group": "Variables for use in worked solution", "definition": "abs(Is)", "description": "", "templateType": "anything", "can_override": false}, "IsPh": {"name": "IsPh", "group": "Variables for use in worked solution", "definition": "arg(Is)*180/pi", "description": "", "templateType": "anything", "can_override": false}, "SsMag": {"name": "SsMag", "group": "Variables for use in worked solution", "definition": "abs(Ss)", "description": "", "templateType": "anything", "can_override": false}, "SsPh": {"name": "SsPh", "group": "Variables for use in worked solution", "definition": " arg(Ss)*180/pi", "description": "", "templateType": "anything", "can_override": false}, "isphtimesminusone": {"name": "isphtimesminusone", "group": "Ungrouped variables", "definition": "abs(isph)*-1", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "abs(Xl-Xc) >20", "maxRuns": 100}, "ungrouped_variables": ["XC1", "Xl1", "Xt", "Rt", "Zt", "Is", "Ss", "Ps", "Qs", "pf", "isphtimesminusone"], "variable_groups": [{"name": "FreqUsedDefinitions", "variables": ["minTol", "maxTol"]}, {"name": "Given in question", "variables": ["Vs", "XL", "XC", "R1"]}, {"name": "Variables for use in worked solution", "variables": ["ZtMag", "ZtPh", "IsMag", "IsPh", "SsMag", "SsPh"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": true, "customName": "Source Current", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The magnitude, $|Is|$ = [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

", "gaps": [{"type": "engineering-answer", "useCustomName": true, "customName": "Is Mag", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "qty(abs(Is),'A')", "right": "2", "close": "5", "C1": "75", "C2": "50", "C3": "25"}}, {"type": "engineering-answer", "useCustomName": true, "customName": "Is Deg", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "qty(arg(Is)*(180/pi),'deg')", "right": "2", "close": "5", "C1": "75", "C2": "50", "C3": "25"}}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": " Total Power", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The total power, (also called apparant power is) is [[0]] at an angle of [[1]]

(Include units in your answer e.g. 99.9 W, 99.9 VA or 99.9 VAr)

", "gaps": [{"type": "engineering-answer", "useCustomName": true, "customName": "Ss Mag", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "qty(abs(Ss),'VA')", "right": "2", "close": "5", "C1": "75", "C2": "50", "C3": "25"}}, {"type": "engineering-answer", "useCustomName": true, "customName": "Ss Deg", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "qty(arg(Ss)*(180/pi),'deg')", "right": "2", "close": "5", "C1": "75", "C2": "50", "C3": "25"}}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": " Active Power", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The active power, (also called real power ) is [[0]]

(Include units in your answer e.g. 99.9 W, 99.9 VA or 99.9 VAr)

", "gaps": [{"type": "engineering-answer", "useCustomName": true, "customName": "Real Power", "marks": "5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "qty(Ps,'W')", "right": "2", "close": "5", "C1": "75", "C2": "50", "C3": "25"}}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": " Reactive Power", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The reactive power, (also called imaginary power) is [[0]] VAr.

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Q noUnit", "marks": "5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "Qs*minTol", "maxValue": "Qs*maxTol", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": "Power factor", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The power factor is [[0]]

Note, power factor is unitless. Give at least four significant figures in your answer.

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "pf", "marks": "5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "pf*minTol", "maxValue": "pf*maxTol", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "2021_Midterm_7_RLCParrallel_VI_v01", "extensions": ["geogebra", "quantities", "weh"], "custom_part_types": [{"source": {"pk": 19, "author": {"name": "William Haynes", "pk": 2530}, "edit_page": "/part_type/19/edit"}, "name": "Engineering Accuracy with units", "short_name": "engineering-answer", "description": "

A value with units marked right if within an adjustable % error of the correct value. Marked close if within a wider margin of error.

Modify the unit portion of the student's answer by

1. replacing \"ohms\" with \"ohm\" case insensitive

2. replacing '-' with ' '

3. replacing '°' with ' deg'

to allow answers like 10 ft-lb and 30°

This fixes the student answer for two common errors.

If student_units are wrong - replace with correct units

If student_scalar has the wrong sign - replace with right sign

If student makes both errors, only one gets fixed.

Only marked close if the student actually has the right sign.

series of R and L and C

find Z (mag and Phase)

step find Xl

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

A circuit consists of a two parrallel branches and is supplied by a {f} Hz, {Vs} V ac voltage source.

Branch 1 $R= $ {qty({R1},'ohm')} and $L = ${qty(L1mH,'mH')}

Branch 2 $R= $ {qty({R2},'ohm')}, and $C = ${qty(C2uF,'uF')}

Find the current flowing from the source. ${Is}$.

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Current in Branch 1, [[0]] at an angle of [[1]]

Give units in your answer, e.g. deg, ohm, A, V

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Current in Branch 2, [[0]] at an angle of [[1]]

Give units in your answer, e.g. deg, ohm, A, V

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The magnitude, $|Is|$ = [[0]] at an angle of [[1]]

Give units in your answer, e.g. deg, ohm, A, V, H or F

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A value with units marked right if within an adjustable % error of the correct value. Marked close if within a wider margin of error.

Modify the unit portion of the student's answer by

1. replacing \"ohms\" with \"ohm\" case insensitive

2. replacing '-' with ' '

3. replacing '°' with ' deg'

to allow answers like 10 ft-lb and 30°

This fixes the student answer for two common errors.

If student_units are wrong - replace with correct units

If student_scalar has the wrong sign - replace with right sign

If student makes both errors, only one gets fixed.

Only marked close if the student actually has the right sign.

series of R and L and C

find Z (mag and Phase)

step find Xl

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

A {Vs} V, 60 Hz ac voltage source supplies a circuit which has a resistance of {Rt} ohms and a reactance of {Xt} ohms connected in series.

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The total power, (also called apparant power is) is [[0]] at an angle of [[1]]

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The active power, (also called real power ) is [[0]]

(Include units in your answer e.g. 99.9 W, 99.9 VA or 99.9 VAr)

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The active power, (also called real power ) is [[0]] VAr

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Qs noUnit", "marks": "5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "Qs*minTol", "maxValue": "Qs*maxTol", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": "Power Factor", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The power factor is [[0]]

Note, power factor is unitless, give 4 significant figures in your answer.

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A value with units marked right if within an adjustable % error of the correct value. Marked close if within a wider margin of error.

Modify the unit portion of the student's answer by

1. replacing \"ohms\" with \"ohm\" case insensitive

2. replacing '-' with ' '

3. replacing '°' with ' deg'

to allow answers like 10 ft-lb and 30°

This fixes the student answer for two common errors.

If student_units are wrong - replace with correct units

If student_scalar has the wrong sign - replace with right sign

If student makes both errors, only one gets fixed.

Only marked close if the student actually has the right sign.

given Voltage and loads calculated all the line currents

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

A 3-phase 4-wire load is supplied by a {VLL} V 3 phase supply.

The load consists of three equal loads of reistance, $R= $ {qty({R1},'ohm')} and reactance $X= $ {qty({X1},'ohm')}.

Reference Phasor: Use the phase voltage between Line1 and Neutral as your zero degree reference phasor.

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The current in line 1, $\\bar{I1}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

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If you are told a voltage from a 3 ph power source, assume it is line-to-line voltage unless told other wise.

So $VLL = {VLL} $

For a balanced 3 phase star source, the line-to-neutral voltage (also called the phase voltage) is smaller than line-to-line voltage.

The ratio is $\\sqrt(3)$

So we have $VLN = VLL/\\sqrt(3) $

So we have $VLN = ${VLN} V

We are told to take V1N as refereence so we have

$V1N = {VLN}$ at $0 deg$
$V2N = {VLN}$ at $-120 deg$
$V1N = {VLN}$ at $120 deg$

"}, {"type": "information", "useCustomName": true, "customName": "Ohms Law", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Now use Ohm's Law for Phase 1

V1N = ${VLN}$ at $0 deg$

$Z1 = {R1} + j{X1}$

$I1 = V1N/Z1 $ ... remember to use phasors/complex numbers.

"}], "gaps": [{"type": "engineering-answer", "useCustomName": true, "customName": "I1 Mag", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "qty(abs(I1),'A')", "right": "2", "close": "5", "C1": "75", "C2": "50", "C3": "25"}}, {"type": "engineering-answer", "useCustomName": true, "customName": "I1 Deg", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "qty(arg(I1)*(180/pi),'deg')", "right": "2", "close": "5", "C1": "75", "C2": "50", "C3": "25"}}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": "I2 Current", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The current in line 2, $\\bar{I2}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

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Now use Ohm's Law for Phase2

V2N = ${VLN}$ at $ -120 deg$

The load is balanced. this means the impedance of each phase is equal.

So we can say Z1=Z2=Z3

Z2 = {R2} + ({X2})j

I2 = V2N/Z2 ... remember to use phasors/complex numbers.

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The current in line 3, $\\bar{I3}$ = [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

Now use Ohm's Law for Phase 3

V3N = ${VLN}$ at $120 deg$

As before , the load is balanced. this means the impedance of each phase is equal.

So we can say Z1=Z2=Z3

Z2 = {R2} + ({X2})j

I2 = V3N/Z2 ... remember to use phasors/complex numbers.

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The current in the neutral line is, $\\bar{I_N}$ = [[0]] Amps.

No need to include the word 'Amps' in your answer.

At the star point of the load examine currnets flowing in and out.

sum of currents OUT = sum of current IN

$ I_N = I1 + I2 + I3 $

$ I_N = {I1} + {I2} + {I3} $

... remember to add complex numbers (i.e. include angles or j-terms)

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A value with units marked right if within an adjustable % error of the correct value. Marked close if within a wider margin of error.

Modify the unit portion of the student's answer by

1. replacing \"ohms\" with \"ohm\" case insensitive

2. replacing '-' with ' '

3. replacing '°' with ' deg'

to allow answers like 10 ft-lb and 30°

This fixes the student answer for two common errors.

If student_units are wrong - replace with correct units

If student_scalar has the wrong sign - replace with right sign

If student makes both errors, only one gets fixed.

Only marked close if the student actually has the right sign.

given Voltage and loads calculated all the line currents

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

A 3-phase 4-wire load is supplied by a {VLL} V 3 phase supply.

The red phase of the star load consists has reistance, $R= $ {qty({R1},'ohm')} and reactance $X= $ {qty({X1},'ohm')}.

The yellow phase of the star load consists has resistance $R= $ {qty({R2},'ohm')} and reactance $X= $ {qty({X2},'ohm')}.

The blue phase of the star load consists has resistance $R= $ {qty({R3},'ohm')} and reactance $X= $ {qty({X3},'ohm')}.

Reference Phasor: Use the phase voltage between Line1 and Neutral as your zero degree reference phasor.

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The current in line 1, $\\bar{I1}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": true, "customName": "Phase voltages", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Remember when a voltage is given describing a 3 phase source, unless you are told otherwise, it is the line-to-line voltage. VLL = {VLL} V

Remember for a star arrangement Line-to-line voltages are bigger than Line-to-neutral by $\\sqrt(3).

So we can say

$V1N = ${VLL}$ /\\sqrt(3)$ at 0 deg = {VLN} at 0 deg
$V2N = ${VLL}$ /\\sqrt(3)$ at -120 deg= {VLN} at -120 deg
$V3N = ${VLL}$ /\\sqrt(3)$ at 120 deg= {VLN} at 120 deg

$V1N = ${V1N} V at 0 deg
$V2N = ${V1N} V at -120 deg
$V3N = ${V1N} V at 120 deg

$Z1 = ${R1} + ({X1})j

$V1N = $ {V1N}

So I1 = V1N/Z1 = ({V1N})/({Z1})

The current in line 2, $\\bar{I2}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

Remember when a voltage is given describing a 3 phase source, unless you are told otherwise, it is the line-to-line voltage. VLL = {VLL} V

Remember for a star arrangement Line-to-line voltages are bigger than Line-to-neutral by $\\sqrt(3).

So we can say

$V1N = {VLL} /\\sqrt(3)$ at 0 deg ... we are told to use this as zero degree reference phasor.
$V2N = {VLL} /\\sqrt(3)$ at -120 deg
$V3N = {VLL} /\\sqrt(3)$ at 120 deg

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$Z2 = ${R2} + ({X2})j

$V2N = $ {V2N}

So I2 = V2N/Z2 = ({V2N})/({Z2})

The current in line 3, $\\bar{I3}$ = [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

$Z3 = ${R3} + ({X3})j

$V3N = $ {V3N}

So I3 = V3N/Z3 = ({V3N})/({Z3})

The current in the neutral line is, $\\bar{I_N}$ = [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

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currents IN = currents OUT ...at the star node in the laod.

$\\bar{I_N}=\\bar{I_1}+\\bar{I_2}+\\bar{I_3}$

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A value with units marked right if within an adjustable % error of the correct value. Marked close if within a wider margin of error.

Modify the unit portion of the student's answer by

1. replacing \"ohms\" with \"ohm\" case insensitive

2. replacing '-' with ' '

3. replacing '°' with ' deg'

to allow answers like 10 ft-lb and 30°

This fixes the student answer for two common errors.

If student_units are wrong - replace with correct units

If student_scalar has the wrong sign - replace with right sign

If student makes both errors, only one gets fixed.

Only marked close if the student actually has the right sign.

", "definition": "right_sign and percent_error <= settings['close']"}, {"name": "right_sign", "description": "", "definition": "sign(student_scalar) = sign(correct_quantity) "}], "settings": [{"name": "correctAnswer", "label": "Correct Quantity.", "help_url": "", "hint": "The correct answer given as a JME quantity.", "input_type": "code", "default_value": "", "evaluate": true}, {"name": "right", "label": "% Accuracy for right.", "help_url": "", "hint": "Question will be considered correct if the scalar part of the student's answer is within this % of correct value.", "input_type": "code", "default_value": "0.2", "evaluate": true}, {"name": "close", "label": "% Accuracy for close.", "help_url": "", "hint": "Question will be considered close if the scalar part of the student's answer is within this % of correct value.", "input_type": "code", "default_value": "1.0", "evaluate": true}, {"name": "C1", "label": "Close with units.", "help_url": "", "hint": "Partial Credit for close value with appropriate units. if correct answer is 100 N and close is ±1%,
99 N is accepted.", "input_type": "percent", "default_value": "75"}, {"name": "C2", "label": "No units or wrong sign", "help_url": "", "hint": "Partial credit for forgetting units or using wrong sign.
If the correct answer is 100 N, both 100 and -100 N are accepted.", "input_type": "percent", "default_value": "50"}, {"name": "C3", "label": "Close, no units.", "help_url": "", "hint": "Partial Credit for close value but forgotten units.
This value would be close if the expected units were provided. If the correct answer is 100 N, and close is ±1%,
99 is accepted.", "input_type": "percent", "default_value": "25"}], "public_availability": "always", "published": true, "extensions": ["quantities"]}], "resources": [["question-resources/Delta2.JPG", "/srv/numbas/media/question-resources/Delta2.JPG"], ["question-resources/delta1.JPG", "/srv/numbas/media/question-resources/delta1.JPG"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "William Haynes", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2530/"}, {"name": "Judy Rea", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/8753/"}], "tags": [], "metadata": {"description": "

given Voltage and loads calculated all the line currents

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

A 3-phase 4-wire load is supplied by a {VLL} V 3 phase supply.

The load consists of three equal loads of reistance, $R= $ {qty({R1},'ohm')} and reactance $X= $ {qty({X1},'ohm')}.

Find the phase currents and the line currents flowing in this network.

Reference Phasor: Use the voltage between Line1 and Line2 as your zero degree reference phasor.

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As instructed in the question the zero degree reference phasor is line to line voltage, V12

$V_{12} = ${VLL }$ \\angle 0 ^{\\circ} $ , $V_{23} = ${VLL }$\\angle $ [[0]]$^{\\circ} $ , $V_{31} = ${VLL }$\\angle $ [[1]]$^{\\circ} $

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The current in phase A, $\\bar{IphA}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

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We are given resistance and reactance so can calculate $Z = R + j (X_L-X_C)

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Before you find the line current you must first fine the phase currents in each phase.

Remember a delta is like a triangle withe points connectd to the \"line\"

Phase A : is between Line 1 and Line 2
Phase B : is between Line 2 and Line 3
Phase C : is between Line 3 and Line 1

Voltages in DELTA Line to Line voltage = phase voltage ... VphA = V12 etc.

Currents in DELTA phase currents are not the same as line currents.

Calculate each phase current first using ohm's law. IphA = VphA/ZphA

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The current in phase A, $\\bar{IphB}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

We are given resistance and reactance so can calculate $Z = R + j (X_L-X_C)

Before you find the line current you must first fine the phase currents in each phase.

Remember a delta is like a triangle withe points connectd to the \"line\"

Phase A : is between Line 1 and Line 2
Phase B : is between Line 2 and Line 3
Phase C : is between Line 3 and Line 1

Voltages in DELTA Line to Line voltage = phase voltage ... VphA = V12 etc.

Currents in DELTA phase currents are not the same as line currents.

Calculate each phase current first using ohm's law. IphA = VphA/ZphA

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The current in phase C, $\\bar{IphC}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

We are given resistance and reactance so can calculate $Z = R + j (X_L-X_C)

Before you find the line current you must first fine the phase currents in each phase.

Remember a delta is like a triangle withe points connectd to the \"line\"

Phase A : is between Line 1 and Line 2
Phase B : is between Line 2 and Line 3
Phase C : is between Line 3 and Line 1

Voltages in DELTA Line to Line voltage = phase voltage ... VphA = V12 etc.

Currents in DELTA phase currents are not the same as line currents.

Calculate each phase current first using ohm's law. IphA = VphA/ZphA

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The current in line 1, $\\bar{I1}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

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The current in line 2, $\\bar{I2}$ is [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

The current in line 3, $\\bar{I3}$ = [[0]] at an angle of [[1]].

(Include units in your answer e.g. 9.99 ohm, 9.99 A, 9.99 V and 99.9 deg)

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given Voltage and loads calculated all the line currents

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A 3-phase 4-wire load is supplied by a {VLL} V 3 phase supply.

The load consists of three loads

Phase A: $R= $ {qty({R1},'ohm')} and reactance $X= $ {qty({X1},'ohm')}.
Phase B: $R= $ {qty({R2},'ohm')} and reactance $X= $ {qty({X2},'ohm')}.
Phase C: $R= $ {qty({R3},'ohm')} and reactance $X= $ {qty({X3},'ohm')}.

Find the line currents flowing in this network.

Reference Phasor: Use the voltage between Line1 and Line2 as your zero degree reference phasor.

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As instructed in the question the zero degree reference phasor is line to line voltage, V12. Fill in the angles of the other two voltages

$V_{12} = ${VLL }$ \\angle 0 ^{\\circ} V$ , $V_{23} = ${VLL }$\\angle $ [[0]]$^{\\circ} V$ , $V_{31} = ${VLL }$\\angle $ [[1]]$^{\\circ} V$

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In DELTA loads, the phase voltages are the same as the line to line voltages so

$V_{A} = ${VLL }$ \\angle 0 ^{\\circ} V$ , $V_{B} = ${VLL }$\\angle $ [[0]]$^{\\circ} V$ , $V_{C} = ${VLL }$\\angle $ [[1]]$^{\\circ} V$

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The impedance in each phase is;

$Z_{A} = ($ [[0]]$ \\angle $[[1]]$ ^{\\circ}) \\Omega$ ,

$Z_{B} = ($ [[2]]$ \\angle $[[3]]$ ^{\\circ}) \\Omega$ ,

$Z_{C} = ($ [[4]]$ \\angle $[[5]]$ ^{\\circ} )\\Omega$

Use $Z = R + j(X_L - X_C) $ for each phase.

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The phase currents in each phase are

$I_{A} = ($ [[0]]$ \\angle $[[1]]$ ^{\\circ} )A$ ,

$I_{B} = ($ [[2]]$ \\angle $[[3]]$ ^{\\circ} )A$ ,

$I_{C} = ($ [[4]]$ \\angle $[[5]]$ ^{\\circ} )A$

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For each phase use Ohms Law

current through a phase = voltage across that phase / impedence ofhta that phase

e.g. $IphA = \\frac{VphA}{ ZphA}$ and so on.

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The Line currents in flowing to this load are

$I_{1} = ( $ [[0]]$ \\angle $[[1]]$ ^{\\circ} )A$ ,

$I_{2} = ($ [[2]]$ \\angle $[[4]]$ ^{\\circ} )A$ ,

$I_{2} = ($ [[3]]$ \\angle $[[5]]$ ^{\\circ} )A$

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given Voltage and loads calculated all the line currents

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A chemical plant consists for 3 loads

Load 1 : {P1} kW at a power factor of {pf1} lagging
Load 2 : An inductive load {P2} kW and an apparant power of {S2}kVA.
Load 3 : {abs(Q3)} kVAr capacitor bank

(a) Calculate the total power factor of the power consumed by these loads in the chemical plant.

(b) Determine the size of capacitor required (in kVAr) to ensure unity power factor.

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Q3 is already negative by definition.

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Complete the follwoing table

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	P	Q	S
Load 1	[[0]] kW	[[3]] kVAr	[[6]] kVA
Load 2	[[1]] kW	[[4]] kVAr	[[7]] kVA
Load 3	[[2]] kW	[[5]] kVAr	\n [[8]] kVA \n

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Use units to help know what info you are given;

kW measures Real or Active Power, P
kW measures Imag or Reactive Power, Q ... be very carefule here on sign of Q.
kW measures Total or Apparant Power, S

If you know any two from these five variables, P, Q, S , pf and $\\theta$ ...

... then you can always find the other values using these two equations

$P^2+Q^2 = |S|^2$
$pf = cos(\\theta) = \\frac{P}{|S|}$

Also, you will need to be careful with polarity of reactive power depeding on leading or lagging loads.

Inductors will have +Q values. Capactitors are the opposite. (i.e. -Q values)

CIVIL ...use this to work if Cap or Inductor.

e.g. current __________ voltage (lead or lag)

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The combined load is there for $($[[0]]}$\\angle $[[1]]$^{\\circ}) kVA = $ [[2]] $kW + j$[[3]] $kVAr$

And the uncorrected power factor is [[4]].

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Add all the real power values to get $P_{total}$

Add all the reactive power values to get $Q_{total}$

Then find total apparent power $S_{total} = P_{total} +jQ_{total}$

Finaly, convert from rectangular to polar form.

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An additional capacative load of [[0]] kVar is required to correct the overall power factor to unity.

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In these questions, we know two things about how the newTotal load will look

New Total Load will have pf=1
New Total Load will have the same real power, P, as before. P = {round(Ptotal)} kW

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	\n P \n [kW] \n	\n Q \n [kVAr] \n	S [kVA]	pf	$\|\\theta\|$
Original Load	{round(Ptotal)}	{round(Qtotal)}
Additional PFC	0
New Total Load = OrigLoad + PFC	{round(Ptotal)}		{round(Ptotal)}	1

Now working on the New Total Row, we know 2 of the 5 variables. So we can calulate the others

(we are really only intersted in Q so we'll only work that one out.)

Since $pf=\\frac{P}{|S|} $, we have |S|= {round(Ptotal)} therefore Q = 0

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	P [kW]	Q [kVAr]	S [kVA]	pf	$\|\\theta\|$
Original Load	{round(Ptotal)}	{round(Qtotal)}
Additional PFC	0
New Total Load = OrigLoad + PFC	{round(Ptotal)}	0	{round(Ptotal)}	1

Final step, looking down the Q column ...

NewTotal Q = Orig Load Q + Qpfc

0 = {round(Qtotal)} + Qpfc

Qpfc = {round(Qtotal*-1)}

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	P [kW]	Q [kVAr]	S [kVA]	pf	$\|\\theta\|$
Original Load	{round(Ptotal)}	{round(Qtotal)}
Additional PFC	0	{round(Qtotal*-1)}
New Total Load = OrigLoad + PFC	{round(Ptotal)}	0	{round(Ptotal)}	1

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An additional capacative load of [[0]] kVar is required to correct the overall power factor to 0.95.

WARNING

I've round the values in this solution to make it readable ... you will need to keep a few decimal places.

JR Nov'21

In these questions, we know two things about how the newTotal load will look

New Total Load will have pf=0.95
New Total Load will have the same real power, P, as before. P = {round(Ptotal)} kW

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	\n P \n [kW] \n	\n Q \n [kVAr] \n	S [kVA]	pf	$\|\\theta\|$
Original Load	{round(Ptotal)}	{round(Qtotal)}
Additional PFC	0
New Total Load = OrigLoad + PFC	{round(Ptotal)}	???	???	0.95

Now working on the New Total Row, we know 2 of the 5 variables. So we can calulate the others

(we are really only intersted in Q so we'll only work that one out.)

Since $pf=\\frac{P}{|S|} $, we have |S|= {round(Ptotal)} * 0.95 therefore Q = {round(QtotalRequired)}

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	P [kW]	Q [kVAr]	S [kVA]	pf	$\|\\theta\|$
Original Load	{round(Ptotal)}	{round(Qtotal)}
Additional PFC	0	????
New Total Load = OrigLoad + PFC	{round(Ptotal)}	{round(QtotalRequired)}	{round(SrequiredMag)}	0.95

Final step, looking down the Q column ...

NewTotal Q = Orig Load Q + Qpfc

{round(QtotalRequired)} = {round(Qtotal)} + Qpfc

Qpfc = {round(QtotalRequired)} - {round(Qtotal)}

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	P [kW]	Q [kVAr]	S [kVA]	pf	$\|\\theta\|$
Original Load	{round(Ptotal)}	{round(Qtotal)}
Additional PFC	0	{round(QtotalRequired)} - {round(Qtotal)}
New Total Load = OrigLoad + PFC	{round(Ptotal)}	{round(QtotalRequired)}	{round(SrequiredMag)}	1

A chemical plant consists for 4 loads

Load 1 : {P1} kW
Load 2 : {S2} kVA at a power factor of {pf2} lagging
Load 3 : {abs(Q3)} kVAr capacitor bank
Load 4 : {S4} kVA and {P4} kVA lagging

What size power factor correction capacitor would you order to correct the overall power factor to 0.95? (Capacitor banks are available in steps of 5 kVAr)

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Q3 is already negative by definition.

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Complete the follwwing table.

Note for S, only the magnitude is required.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	P	Q	\|S\|
Load 1	[[0]] kW	[[4]] kVAr	[[8]] kVA
Load 2	[[1]] kW	[[5]] kVAr	[[9]] kVA
Load 3	[[2]] kW	[[6]] kVAr	\n [[10]] kVA \n
Load 4	[[3]] kW	[[7]] kVAr	\n [[11]] kVA \n

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Given any two from $P,Q,S,pf$ the rest of the values can be found by various manipulation of the following equations.

$P^2 + Q^2 = |S|^2$

$ pf = \\frac{P}{|S|} = cos(\\Theta) $

Also, you will need to be careful with polarity of reactive power $Q$ depeding on leading or lagging loads.

CIVIL - helps to tell if the load is capacitive or inductive.

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The combined load is there for $($[[0]]}$\\angle $[[1]]$^{\\circ}) kVA = $ [[2]] $kW + j$[[3]] $kVAr$

And the uncorrected power factor is [[4]].

Add all the real power values to get $P_{total}$

Add all the reactive power values to get $Q_{total}$

Then find total apparent power $S_{total} = P_{total} +jQ_{total}$

Finaly, convert from rectangular to polar form.

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An additional capacative load of [[0]] kVar is required to correct the overall power factor to 0.95.

The power rating of the new capacitor should be set to [[1]] kVAr.

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To get unity power factor you would need to fully compensate for the $Q_{total}$ so a capacitor sized $-Q_{total}$ would do the job.

Here we are asked to correct the power factor to 0.95. Adding a PFC cap load will not change the real power of the original load.

On the row for New Total we know P and pf (=0.95) we can find $Q_{newTotal}$.

Final step then is to $Q_{origLoad}$ + $Q_{pfc}$ = $Q_{newTotal}$ and solve for $Q_{pfc}.$

Then round to the nearest 5 kVAr.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	P	Q
Uncorrected Load	{round(Ptotal)}	{round(Qtotal)}
Additional Cap (PFC)	0 (no real power consumbed by PFC)	$Q_{pfc}$
Corrected load = Uncorrected + PFC	\n {round(Ptotal)} + 0 = {round(Ptotal)} \n

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given Vdc, Iop , Top, res find L.

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An circuit consists of a {Rkohm} k\u2126 resistor, a {CuF} uF capacitor and a switch, all connected in series.

It is supplied by a DC voltage source, {Vdc} V.

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Initially, immediately after the switch is closed the current flowing is [[1]] mA.

After some time, when all the transients have died out, the current flowing is [[0]] mA.

The time constant for this circuit is [[2]] ms.

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What initial current,$I_{0}$, immediatly after the switch is closed?

Initially, the cap is uncharged so no voltage is dropped across it meaning the full source voltage is dropped across the resister, hence, by ohm's law, $I_{0} = \\frac{V_{DC}}{R} $.

What is the steady state current,$I_{SS}$,will be reached when all the transients settle down?

This is zero, since when all transients are settled down capacitors are like open circuits.

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$\\tau=RC$

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After {t1*1000} ms, the current is [[0]] mA.

After {t2*1000} ms, the current is [[1]] mA.

After [[2]] ms for the current is {round(I3*1000)} mA.

After [[3]] ms for the current is {round(I4*1000)} mA.

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So the equation descibing the transient current in this circuit is

$i(t) = I_{0}(e^{\\frac{-t}{\\tau}})$ where ${\\tau} = {{RC}}$.

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given Vdc, Iop , Top, res find L.

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An circuit consists of a {R} \u2126 resistor, a {L} H n inductor and a switch, all connected in series. It is supplied by a DC voltage source, {Vdc} V.

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Initally, immediately after the switch is closed, how much current flows? [[0]] mA

After the switch has closed and all the transients have died out, the current flowing is [[1]] mA.

The time constant for this circuit is [[2]] ms.

What steady state current,$I_{SS}$,will be reached when all the transients settle down?

When all transients are settled down inductors appear like short circuits and capacitors are like open circuits.

So in steady state, by ohm's law, $I_{SS} = \\frac{V_{DC}}{R} $

$\\tau=\\frac L R$

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After {t1*1000} ms, the current is [[0]] mA.

After {t2*1000} ms, the current is [[1]] mA.

It takes [[2]] ms for the current to grow to {round(I3*1000)} mA.

It takes [[3]] ms for the current to grow to {round(I4*1000)} mA.

So the equation descibing the transient current in this circuit is

$i(t) = I_{SS}(1-e^{\\frac{-t}{\\tau}})$ where ${\\tau} = {\\frac{L}{R}}$.

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given Vdc, Iop , Top, res find L.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

A relay coil has a resistance of {R} \u2126 and the current required to operate the relay is {{Iop}*1000} mA.

When the coil is connected to a {Vdc} V d.c. power supply the operating time between, switch-on and the start of the relay operation is {top*1000} ms.

Find L requried to achieve this behaviour.

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operating time in seconds (not Milli

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current in A

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Use iop = Iss(1-exp^-top/TAU)

ln(1-iop/Iss) = -top/TAU

TAU = -top/ln(1-iop/Iss)

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To get the relay to operate after {top*1000} ms an inductor of [[0]] H is needed.

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The message has been 'relayed' from primary circuit to the secondary circuit with a know time-delay.

The relay is supplied a by a dc voltage source and consists of an inductor and a resistor in series.

When current flows through the coil of the relay circuit (BLUE), the magnetic field is created(RED). When the current reaches the operating current {Iop}, then the relay \"operates\" because the magnetic field is now strong enought to cause the switch in secondary circuit to close (GREEN).

In this question, we are asked to design (i.e. select) a value of L that will set a {top*1000} ms time-delay between closing the control switch (BLUE) to when the secondary switch will close. (GREEN)

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What steady state current,$I_{SS}$,will be reached when all the transients settle down?

When all transients are settled down inductors appear like short circuits and capacitors are like open circuits.

So in steady state, by ohm's law, $I_{SS} = \\frac{V_{DC}}{R} = ${Iop} A = {Iop*1000} mA

We would like the current have grown to the operating current, {Iop*1000} mA after {top*1000} ms.

Currents grow in an RL circuit according the equation $i(t) = I_{SS}(1-e^{\\frac{-t}{\\tau}})$

So fill ... $i(t) = i(${top}$) = ${Iop} , $t = ${top}, $I_{SS} = ${Iss} into the equation and rearrange to fine $\\tau$.

Rearranged equation to calculate

${\\tau}=\\frac{-{t_{op}}}{ln(1 -\\frac{i_{op}}{I_{SS}})} $

So ${\\tau}$ is ...

${\\tau} = \\frac{L}{R}$

Since R={R}, and we now know ${\\tau}$ then we can find L.

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A value with units marked right if within an adjustable % error of the correct value. Marked close if within a wider margin of error.

Modify the unit portion of the student's answer by

1. replacing \"ohms\" with \"ohm\" case insensitive

2. replacing '-' with ' '

3. replacing '°' with ' deg'

to allow answers like 10 ft-lb and 30°

This fixes the student answer for two common errors.

If student_units are wrong - replace with correct units

If student_scalar has the wrong sign - replace with right sign

If student makes both errors, only one gets fixed.

Only marked close if the student actually has the right sign.

", "definition": "right_sign and percent_error <= settings['close']"}, {"name": "right_sign", "description": "", "definition": "sign(student_scalar) = sign(correct_quantity) "}], "settings": [{"name": "correctAnswer", "label": "Correct Quantity.", "help_url": "", "hint": "The correct answer given as a JME quantity.", "input_type": "code", "default_value": "", "evaluate": true}, {"name": "right", "label": "% Accuracy for right.", "help_url": "", "hint": "Question will be considered correct if the scalar part of the student's answer is within this % of correct value.", "input_type": "code", "default_value": "0.2", "evaluate": true}, {"name": "close", "label": "% Accuracy for close.", "help_url": "", "hint": "Question will be considered close if the scalar part of the student's answer is within this % of correct value.", "input_type": "code", "default_value": "1.0", "evaluate": true}, {"name": "C1", "label": "Close with units.", "help_url": "", "hint": "Partial Credit for close value with appropriate units. if correct answer is 100 N and close is ±1%,
99 N is accepted.", "input_type": "percent", "default_value": "75"}, {"name": "C2", "label": "No units or wrong sign", "help_url": "", "hint": "Partial credit for forgetting units or using wrong sign.
If the correct answer is 100 N, both 100 and -100 N are accepted.", "input_type": "percent", "default_value": "50"}, {"name": "C3", "label": "Close, no units.", "help_url": "", "hint": "Partial Credit for close value but forgotten units.
This value would be close if the expected units were provided. If the correct answer is 100 N, and close is ±1%,
99 is accepted.", "input_type": "percent", "default_value": "25"}], "public_availability": "always", "published": true, "extensions": ["quantities"]}], "resources": [["question-resources/Delta2.JPG", "/srv/numbas/media/question-resources/Delta2.JPG"], ["question-resources/delta1.JPG", "/srv/numbas/media/question-resources/delta1.JPG"], ["question-resources/Capture_OVvVQTf.JPG", "/srv/numbas/media/question-resources/Capture_OVvVQTf.JPG"], ["question-resources/RLcct.png", "/srv/numbas/media/question-resources/RLcct.png"], ["question-resources/rcNetwork.JPG", "/srv/numbas/media/question-resources/rcNetwork.JPG"]]}