","comment":{"@type":"Comment","text":"To calculate the maximum/peak current and impedance, first determine the current flowing through the lamp, which is the rms current."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.25\\ A$$, $$80\\Omega $$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.25\\ A$$, $$56\\Omega $$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.3535\\ A$$, $$80\\Omega $$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.3535\\ A$$, $$800\\Omega $$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{We are given}\\,{{\\epsilon }_{rms}}=200\\ V,P=5\\ W\\,\\text{and}\\,V=20\\ V.\\\\ \\text{The current flowing through the lamp will be}\\,I=\\dfrac{P}{V}=\\dfrac{5}{20}=0.25\\ A\\\\\\text{This is nothing but the rms current, i.e.,}\\,{{I}_{rms}}=0.25\\ A\\\\\\text{The maximum value of the current will be}\\\\ {{I}_{max}}=\\sqrt{2}{{I}_{rms}}\\\\ \\Rightarrow{{I}_{max}} =\\sqrt{2}\\times 0.25\\\\ \\Rightarrow{{I}_{max}} =0.3535\\ A\\\\\\text{The impedance of the circuit will be:}\\\\ Z=\\dfrac{{{\\epsilon }_{rms}}}{{{I}_{rms}}}\\\\ \\Rightarrow Z =\\dfrac{200}{0.25}\\\\\\therefore Z =800\\Omega$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Direct and alternating current Quiz 1","text":"Consider the following oscillating circuit that is used to tune the radio frequency of an AM/FM radio. The voltage drops across the current elements are $${{V}_{L}}=50\\ V$$, $${{V}_{C}}=80\\ V$$ and $${{V}_{R}}=40\\ V$$ respectively. What is the voltage of the alternator used? ","comment":{"@type":"Comment","text":"Sketch out a phasor diagram to determine the resultant potential difference which is nothing but the alternator voltage. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$30\\ V$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$40\\ V$$ ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$90\\ V$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$50\\ V$$","position":2,"answerExplanation":{"@type":"Comment","text":"$$\\text{We are given}\\,|\\overrightarrow{{{V}_{L}}}|=50\\ V,|\\overrightarrow{{{V}_{C}}}|=80\\ V\\,\\text{and}\\,|\\overrightarrow{{{V}_{R}}}|=40\\ V.\\\\\\text{The phase difference between the potential difference across the inductor and the capacitor is}\\,\\pi\\,\\text{since they are vectorially directed opposite to each other.}\\\\ \\text{The resultant potential difference across the combination of inductor and capacitor will be:}\\\\ |\\overrightarrow{{{V}^{\\prime }}}|=|\\overrightarrow{{{V}_{C}}}|-|\\overrightarrow{{{V}_{L}}}|\\\\ \\Rightarrow |\\overrightarrow{{{V}^{\\prime }}}|=80-50=30\\ V\\\\ \\text{The phase difference between the potential difference across the inductor-capacitor combo and the resistor is}\\,\\dfrac{\\pi }{2}\\,\\text{since they are directed perpendicular to each other.}\\\\\\text{The resultant potential difference is nothing but the voltage of the alternator and is given as:}\\\\ {{\\vec{V}}_{alternator}}=\\sqrt{|\\overrightarrow{{{V}^{\\prime }}}{{|}^{2}}+|\\overrightarrow{{{V}_{R}}}{{|}^{2}}+2|\\overrightarrow{{{V}^{\\prime }}}||\\overrightarrow{{{V}_{R}}}|cos\\left( \\dfrac{\\pi }{2} \\right)}\\\\ \\Rightarrow{{\\vec{V}}_{alternator}} =\\sqrt{{{30}^{2}}+{{40}^{2}}}\\\\ \\therefore{{\\vec{V}}_{alternator}} =50\\ V\\\\ \\text{The phasor diagram looks as follows.}$$
","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Direct and alternating current Quiz 1","text":"The admittance of a series $$LR$$ circuit is found to be $$2\\times {{10}^{-3}}\\ {{\\Omega }^{-1}}$$.Determine the angular frequency of oscillation of this circuit, given $$L=0.5\\ H$$ and $$R=100\\Omega $$.","comment":{"@type":"Comment","text":"Use the expression for total impedance and expand the inductive reactance to find the angular frequency. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$150\\ {{s}^{-1}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$490\\ {{s}^{-1}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$98\\ {{s}^{-1}}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$980\\ {{s}^{-1}}$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$\\text{We are given that}\\,R=100\\Omega\\,\\text{and}\\,L=0.5\\ H.\\\\ \\text{The admittance of this circuit}\\,Y=2\\times {{10}^{-3}}\\ {{\\Omega }^{-1}}.\\\\ \\text{The admittance is the inverse of impedance, i.e.,}\\,Y=\\dfrac{1}{Z}.\\\\\\text{The impedance of this circuit can be given as:}\\\\ Z=\\sqrt{{{R}^{2}}+X_{L}^{2}}\\\\\\text{We also know that}\\,{{X}_{L}}=L\\omega\\\\\\Rightarrow \\dfrac{1}{Y}=\\sqrt{{{R}^{2}}+{{L}^{2}}{{\\omega }^{2}}}\\\\\\Rightarrow \\omega =\\dfrac{\\sqrt{1-{{(2\\times {{10}^{-3}})}^{2}}({{100}^{2}})}}{2\\times {{10}^{-3}}\\times 0.5}\\\\\\Rightarrow\\omega=\\dfrac{\\sqrt{1-0.04}}{{{10}^{-3}}}\\\\\\Rightarrow \\omega =\\sqrt{0.96}\\times {{10}^{3}}\\\\\\therefore \\omega = 979.79\\ {{s}^{-1}}\\approx 980\\ {{s}^{-1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Direct and alternating current Quiz 1","text":"When a charged capacitor is connected to an inductor, both the current and the charge on the capacitor oscillate. In the given figure, a capacitor is initially charged when $${{K}_{1}}$$ is open and $${{K}_{2}}$$ is closed. Then $${{K}_{1}}$$ is closed and $${{K}_{2}}$$ is opened such that the capacitor is shorted across the inductor. Determine $$\\\\$$ i) The maximum value of charge on the capacitor. $$\\\\$$ ii) The maximum value of current in the circuit. $$\\\\$$ iii) What would be the resultant equation of current as a function of time?
","comment":{"@type":"Comment","text":"When the capacitor is fully charged and not discharged, it has the highest charge. To determine the maximum current, first determine the oscillation frequency."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1.08\\times {{10}^{-10}}\\ C$$, $$6.79\\times {{10}^{-4}}\\ A$$ and $$-(6.79\\times {{10}^{-4}})\\ cos(\\omega t)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.75\\times {{10}^{-12}}\\ C$$, $$4.72\\times {{10}^{-6}}\\ A$$ and $$-(4.72\\times {{10}^{-6}})\\ sin(\\omega t)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.75\\times {{10}^{-12}}\\ C$$, $$4.72\\times {{10}^{-6}}\\ A$$ and $$-(4.72\\times {{10}^{-6}})\\ cos(\\omega t)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1.08\\times {{10}^{-10}}\\ C$$, $$6.79\\times {{10}^{-4}}\\ A$$ and $$-(6.79\\times {{10}^{-4}})\\ sin(\\omega t)$$","position":0,"answerExplanation":{"@type":"Comment","text":"$$\\text{From the figure, we have}\\,V=12\\ V, C=9\\times {{10}^{-12}}\\ F\\,\\text{and}\\,L=2.81\\times {{10}^{-3}}\\ H.\\\\\\text{The oscillation of charge can be given as}\\,Q={{Q}_{max}}\\ cos(\\omega t)\\\\\\text{The maximum charge on the capacitor will be the charge it attains when}\\,{{K}_{1}}\\,\\text{is closed and is initially charged, i.e.,}\\\\ {{Q}_{max}}=CV\\\\ \\Rightarrow{{Q}_{max}} =(9\\times {{10}^{-12}})\\times 12\\\\ \\Rightarrow{{Q}_{max}} =1.08\\times {{10}^{-10}}\\ C\\\\\\text{Now, we know that current can be written as}\\\\ I =\\dfrac{dQ}{dt}\\\\ \\Rightarrow I =\\dfrac{d}{dt}({{Q}_{max}}\\ cos(\\omega t))\\\\ \\Rightarrow I =-\\omega {{Q}_{max}}\\ sin(\\omega t)\\\\\\text{The current is maximum when}\\,sin(\\omega t)=-1:\\\\ \\Rightarrow {{I}_{max}}=\\omega {{Q}_{max}}\\\\ \\omega =\\dfrac{1}{\\sqrt{LC}}\\\\ \\Rightarrow\\omega =\\dfrac{1}{\\sqrt{2.81\\times {{10}^{-3}}\\times 9\\times {{10}^{-12}}}}\\\\ \\Rightarrow\\omega =6.288\\times {{10}^{6}}\\ {{s}^{-1}}\\\\\\Rightarrow {{I}_{max}}=(6.288\\times {{10}^{6}})\\times (1.08\\times {{10}^{-10}})\\\\ \\therefore{{I}_{max}} = 6.79\\times {{10}^{-4}}\\ A\\\\\\text{We have}\\,I=-{{I}_{max}}sin(\\omega t)\\\\\\text{Therefore, the resultant equation of current as a function of time will be:}\\\\\\therefore I=-(6.79\\times {{10}^{-4}})\\ sin(\\omega t)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}