// Numbas version: finer_feedback_settings
{"name": "Test SimEq Worksheet", "metadata": {"description": "", "licence": "All rights reserved"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": [""], "variable_overrides": [[]], "questions": [{"name": "Simultaneous Equation - Kirchhoff's Law", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Graeme Woodhall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/30583/"}], "tags": [], "metadata": {"description": "<p>BTEC Unit 3 Maths - Simultaneous Equations</p>", "licence": "All rights reserved"}, "statement": "<p>When Kirchhoff&rsquo;s laws are applied to the two-loop circuit below, <br/>the following simultaneous equations arise (currents in amperes):</p>\n<p>\\[<br/>\\begin{aligned}<br/>(R_1 + R_3)\\,I_{1} \\;-\\; R_3\\,I_{2} &amp;= V_1\\\\[6pt]<br/>-\\;R_3\\,I_{1} \\;+\\; (R_2 + R_3)\\,I_{2} &amp;= V_2<br/>\\end{aligned}<br/>\\]</p>\n<p>with;</p>\n<p>\\( R_1 = \\var{R1}\\,\\Omega,\\;<br/>R_2 = \\var{R2}\\,\\Omega,\\;<br/>R_3 = \\var{R3}\\,\\Omega,\\quad<br/>V_1 = \\var{V1}\\,\\text{V},\\;<br/>V_2 = \\var{V2}\\,\\text{V}. \\)</p>\n<p>Calculate \\(I_{1}\\) and \\(I_{2}\\) in amperes, showing all working.</p>", "advice": "<p style=\"page-break-before: always;\"></p>\n<p><strong>Worked solution &ndash; substitution method</strong></p>\n<p>We start from the symbolic loop equations:</p>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n\\simplify{-R3+(R2+R3)*(R1+R3)/R3}\\,I_{1}\n- \\simplify{(R2+R3)*V1/R3} &= \\var{V2}\n\\end{aligned}\n\n<hr/>\n<h3>1. Insert the numerical values</h3>\n<p>Equation (1&prime;):</p>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n(\\var{R1}+\\var{R3})\\,I_{1} - \\var{R3}\\,I_{2} &= \\var{V1}\n\\end{aligned}\n\n<p>Equation (2&prime;):</p>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n-\\var{R3}\\,I_{1} + (\\var{R2}+\\var{R3})\\,I_{2} &= \\var{V2}\n\\end{aligned}\n\n<hr/>\n<h3>2. Rearrange (1&prime;) to make <em>I<sub>2</sub></em> the subject</h3>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n\\var{R3}\\,I_{2} &= (\\var{R1}+\\var{R3})\\,I_{1} - \\var{V1} \\\\\nI_{2} &= \\frac{(\\var{R1}+\\var{R3})}{\\var{R3}}\\,I_{1} - \\frac{\\var{V1}}{\\var{R3}}\n\\end{aligned}\n\n<hr/>\n<h3>3. Substitute (3) into equation (2&prime;) and simplify</h3>\n<p>Substitution:</p>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n-\\var{R3}\\,I_{1} + (\\var{R2}+\\var{R3})\\left(\\frac{(\\var{R1}+\\var{R3})}{\\var{R3}}\\,I_{1} - \\frac{\\var{V1}}{\\var{R3}}\\right) &= \\var{V2}\n\\end{aligned}\n\n<p>After simplifying the left-hand side:</p>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n\\simplify{-3+(5+3)*(5+3)/3}\\,I_{1}\n- \\simplify{(5+3)*(-27)/3} &= \\var{V2}\n\\end{aligned}\n\n<hr/><hr/>\n<h3>4. Solve for <em>I<sub>1</sub></em></h3>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\nI_{1} &= \\var{I_1}\\;\\text{A}\n\\end{aligned}\n\n<hr/>\n<h3>5. Find <em>I<sub>2</sub></em> using (3)</h3>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\nI_{2} &= \\frac{(\\var{R1}+\\var{R3})}{\\var{R3}}\\,(\\var{I_1}) - \\frac{\\var{V1}}{\\var{R3}} \\\\\n&= \\var{I_2}\\;\\text{A}\n\\end{aligned}\n\n<hr/>\n<h3>6. Check (optional but good practice)</h3>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n- \\var{R3}\\times\\var{I_1} + (\\var{R2}+\\var{R3})\\times\\var{I_2} &= \\var{V2}\\quad\\checkmark\n\\end{aligned}\n\n<hr/>\n<h3>Result</h3>\n<script type=\"math/tex; mode=display\"></script>\n\\begin{aligned}\n\\boxed{\\,I_{1}=\\var{I_1}\\;\\text{A},\\quad I_{2}=\\var{I_2}\\;\\text{A}\\,}\n\\end{aligned}\n\n<p><em>The negative sign on <em>I<sub>2</sub></em> simply means the real current flows opposite to the assumed loop direction.</em></p>", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"I_1": {"name": "I_1", "group": "Ungrouped variables", "definition": "random(-5..5 except 0)\n", "description": "", "templateType": "anything", "can_override": false}, "I_2": {"name": "I_2", "group": "Ungrouped variables", "definition": "random(-5..5 except 0)\n", "description": "", "templateType": "anything", "can_override": false}, "R1": {"name": "R1", "group": "Ungrouped variables", "definition": "random(2,3,4,5,6,8,10)\n", "description": "", "templateType": "anything", "can_override": false}, "R2": {"name": "R2", "group": "Ungrouped variables", "definition": "random(1,2,3,4,5,6,8)\n\n\n", "description": "", "templateType": "anything", "can_override": false}, "R3": {"name": "R3", "group": "Ungrouped variables", "definition": "random(1,2,3,4,5)\n", "description": "", "templateType": "anything", "can_override": false}, "V1": {"name": "V1", "group": "Ungrouped variables", "definition": "(R1+R3)*I_1 - 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