{"id":24813,"date":"2020-02-29T17:20:53","date_gmt":"2020-02-29T11:50:53","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=24813"},"modified":"2021-07-02T15:24:02","modified_gmt":"2021-07-02T09:54:02","slug":"2nd-puc-electronics-previous-year-question-paper-march-2017","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-electronics-previous-year-question-paper-march-2017\/","title":{"rendered":"2nd PUC Electronics Previous Year Question Paper March 2017"},"content":{"rendered":"<p>Students can Download 2nd PUC Electronics Previous Year Question Paper March 2017, Karnataka <a href=\"https:\/\/kseebsolutions.guru\/2nd-puc-electronics-model-question-papers-with-answers\/\">2nd PUC Electronics Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n<h2>Karnataka 2nd PUC Electronics Previous Year Question Paper March 2017<\/h2>\n<h3>2nd PUC Electronics Previous Year Question Paper March 2017<\/h3>\n<p>Time : 3 hrs. 15 min.<br \/>\nMax Marks: 70<\/p>\n<p>Instructions:<\/p>\n<ul>\n<li>The question has five Parts A, B, C and D.<\/li>\n<li>Part &#8211; A has no choice<\/li>\n<li>Part &#8211; D has two parts. Part -1 is from problems, Part &#8211; II is of essay type questions.<\/li>\n<li>Circuit diagram\/timing diagram\/truth tables are drawn wherever necessary.<\/li>\n<li>Problems without necessary formula\/formulae carry no marks.<\/li>\n<\/ul>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; A<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer all questions: 10 x 1 = 10<\/span><\/p>\n<p>Question 1.<br \/>\nWrite the function of source in JFET.<br \/>\nAnswer:<br \/>\nSource supplies charge carriers for current conduction.<\/p>\n<p>Question 2.<br \/>\nName the power amplifier in which conduction angle is 360\u00b0.<br \/>\nAnswer:<br \/>\nClass A.<\/p>\n<p>Question 3.<br \/>\nDefine slew rate.<br \/>\nAnswer:<br \/>\nSlew rate is the rate of change of output.<\/p>\n<p>Question 4.<br \/>\nWhat is the frequency of modulating signal if the bandwidth of AM wave is 10 KHz?<br \/>\nAnswer:<br \/>\n\\(f_{\\mathrm{m}}=\\frac{\\mathrm{BW}}{2}=\\frac{10 \\mathrm{KHz}}{2}=5 \\mathrm{KHz}\\)<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 5.<br \/>\nDefine demodulation.<br \/>\nAnswer:<br \/>\nDemodulation is the process of recovering original modulating signal from the modulated wave.<\/p>\n<p>Question 6.<br \/>\nDraw the circuit symbol of IGBT.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57790\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-1.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 1\" width=\"229\" height=\"86\" \/><\/p>\n<p>Question 7.<br \/>\nWhat is a counter?<br \/>\nAnswer:<br \/>\nCounter is a logic circuit for counting the pulses.<\/p>\n<p>Question 8.<br \/>\nConvert (1101)2 into gray code.<br \/>\nAnswer:<br \/>\n(1011)<sub>G<\/sub><\/p>\n<p>Question 9.<br \/>\nHow many interrupt sources are there in 8051 microcontroller?<br \/>\nAnswer:<br \/>\n6 including RESET or 5 excluding RESET.<\/p>\n<p>Question 10.<br \/>\nWrite C equivalent expression for the mathematical expression \\(\\sqrt{a^{2}+b^{2}}\\)<sup><br \/>\n<\/sup>Answer:<br \/>\nsqrt (a*a+ b*b)<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; B<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any five questions.<\/span><\/p>\n<p>Question 11.<br \/>\nWrite any two differences between FET and BJT.<br \/>\nAnswer:<\/p>\n<table border=\"2px\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"46\"><strong>S.No.<\/strong><\/td>\n<td style=\"text-align: center;\" width=\"245\"><strong>BJT<\/strong><\/td>\n<td style=\"text-align: center;\" width=\"258\"><strong>JFET<\/strong><\/td>\n<\/tr>\n<tr>\n<td width=\"46\">1.<\/td>\n<td width=\"245\">It is a bipolar device.<\/td>\n<td width=\"258\">It is a unipolar device.<\/td>\n<\/tr>\n<tr>\n<td width=\"46\">2.<\/td>\n<td width=\"245\">It is a current controlled device.<\/td>\n<td width=\"258\">It is a voltage controlled device.<\/td>\n<\/tr>\n<tr>\n<td width=\"46\">3.<\/td>\n<td width=\"245\">Current conduction is by both holes and electrons.<\/td>\n<td width=\"258\">Current conduction is by holes or electrons.<\/td>\n<\/tr>\n<tr>\n<td width=\"46\">4.<\/td>\n<td width=\"245\">Input resistance is low.<\/td>\n<td width=\"258\">Input resistance is very high. (MCI)<\/td>\n<\/tr>\n<tr>\n<td width=\"46\">5.<\/td>\n<td width=\"245\">BJT is more noisy.<\/td>\n<td width=\"258\">JFET is less noisy.<\/td>\n<\/tr>\n<tr>\n<td width=\"46\">6.<\/td>\n<td width=\"245\">Switching speed is less.<\/td>\n<td width=\"258\">Switching speed is high.<\/td>\n<\/tr>\n<tr>\n<td width=\"46\">7.<\/td>\n<td width=\"245\">Fabrication in IC is difficult.<\/td>\n<td width=\"258\">Fabrication in IC is simpler.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 12.<br \/>\nMention the steps involved to obtain DC equivalent circuit of an amplifier.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Reduce all the AC source to zero.<\/li>\n<li>open all the capacitors.<\/li>\n<\/ul>\n<p>Question 13.<br \/>\nDistinguish between positive feedback and negative feedback.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Positive feedback is the process of applying a portion of output in phase with the input.<\/li>\n<li>Negative feedback is the process of applying a portion of output out of phase with the input.<\/li>\n<\/ul>\n<p>Question 14.<br \/>\nWrite any two advantages of RC oscillators over LC oscillators.<br \/>\nAnswer:<\/p>\n<ul>\n<li>RC oscillators provide constant output and good stability.<\/li>\n<li>RC oscillators do not require any inductor and hence size and cost get reduced.<\/li>\n<\/ul>\n<p>Question 15.<br \/>\nWrite Shockley&#8217;s diode equation for current through the power diode and explain its terminology.<br \/>\nAnswer:<br \/>\n\\(I=I_{ s }\\left( e^{ \\frac { qv }{ KT } }-1 \\right)\\)<\/p>\n<ul>\n<li>I<sub>s <\/sub>= Reverse saturation current<br \/>\n<a href=\"#_ftnref3\" name=\"_ftn3\"><\/a><\/li>\n<li>T = Absolute temperature<br \/>\n<a href=\"#_ftnref2\" name=\"_ftn2\"><\/a><a href=\"#_ftnref4\" name=\"_ftn4\"><\/a><\/li>\n<li>K = Boltzmann constant<\/li>\n<li>q = charge of electron<\/li>\n<li>V = voltage across diode.<\/li>\n<\/ul>\n<p>Question 16.<br \/>\nDetermine average value of DC from chopper. Given T= 2 ms, T<sub>QN<\/sub>= 0.5 ms and supply voltage is 24V.<br \/>\nAnswer:<br \/>\nDuty ratio = \\( \\frac{T_{o N}}{T}=\\frac{0.5}{2}=0.25\\)<br \/>\nV<sub>dc<\/sub> = Duty ratio x V<sub>s<\/sub> = 0.25 x 24 = 6V.<\/p>\n<p>Question 17.<br \/>\nRealise NOT and OR-gate using NOR-gate<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57789\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-2.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 2\" width=\"666\" height=\"107\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-2.png 666w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-2-300x48.png 300w\" sizes=\"(max-width: 666px) 100vw, 666px\" \/><\/p>\n<p>Question 18.<br \/>\nDistinguish between uplink and downlink signals<br \/>\nAnswer:<\/p>\n<table border=\"2PX\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"205\"><strong>Uplink<\/strong><\/td>\n<td style=\"text-align: center;\" width=\"205\"><strong>Downlink<\/strong><\/td>\n<\/tr>\n<tr>\n<td width=\"205\">1)\u00a0 Signal transmitted from ground station to satellite<br \/>\n2) Frequency range is 6 GHz<\/td>\n<td width=\"205\">1) Signal transmitted from satellite to ground station.<br \/>\n2) Frequency range is 4 GHz.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; C<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any five questions.\u00a0 5 x 3 = 15<\/span><\/p>\n<p>Question 19.<br \/>\nWhat is a DC load line? Mention any two advantages of voltage divider bias.<br \/>\nAnswer:<br \/>\nOperating point is the zero signal values of collector I<sub>c<\/sub> and collector to emitter voltage V<sub>CE <\/sub>of a transistor. DC load line is a straight line drawn on the output characteristics representing output DC voltage and corresponding direct current under no signal conditions.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017\" width=\"172\" height=\"16\" \/><\/p>\n<p>The operating point does not depend upon the value of \u03b2 of the transistor. Hence the operating point does not change its position due to rise in temperature or replacing a transistor of different p value. Hence voltage divider bias provides excellent stabilisation and is preferred to other biasing methods.<\/p>\n<p>Question 20.<br \/>\nCalculate the input impedance of a negative feedback amplifier if input impedance without feedback is 10 KQ. Given A = 100 and \u03b2 = 0.01.<br \/>\nAnswer:<br \/>\nZ<sub>if<\/sub> = Z<sub>i<\/sub>(1 +A\u03b2)<br \/>\n= 10 x 10<sup>3<\/sup> (1 + 100 x 0.01)<br \/>\n= 20 x 10<sup>3<\/sup> \u03a9<br \/>\n= 20k\u03a9<\/p>\n<p>Question 21.<br \/>\nWith block diagram, explain basic communication system.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57788\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-3.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 3\" width=\"613\" height=\"134\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-3.png 613w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-3-300x66.png 300w\" sizes=\"(max-width: 613px) 100vw, 613px\" \/><\/p>\n<p>Question 22.<br \/>\nDraw the equivalent circuit of transmission lines for low frequency. Mention any two types of antennas.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57787\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-4.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 4\" width=\"523\" height=\"131\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-4.png 523w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-4-300x75.png 300w\" sizes=\"(max-width: 523px) 100vw, 523px\" \/><br \/>\nTypes of Antenna:<\/p>\n<ul>\n<li>Helical antenna<\/li>\n<li>Yagi antenna<\/li>\n<li>Loop antenna<\/li>\n<\/ul>\n<p>Question 23.<br \/>\nWrite the circuit diagram, input and output waveforms of SCR halfwave rectifier with RC triggering circuit.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57786\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-5.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 5\" width=\"514\" height=\"216\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-5.png 514w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-5-300x126.png 300w\" sizes=\"(max-width: 514px) 100vw, 514px\" \/><br \/>\nThe SCR is connected with a diode bridge rectifier. The rectifier circuit converts AC to a unidirectional current through the thyristor. The voltage on the capacitor follows the diode rectifier waveform because RC circuit is connected to diode rectifier. During each half cycle when the voltage across the capacitor reaches trigger level which is just enough to supply gate trigger current to thyristor, then thyristor starts conducting. Firing angle \\(\\alpha=\\left(\\frac{x}{\\mathrm{y}}\\right)_{180}\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57785\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-6.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 6\" width=\"205\" height=\"89\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 24.<br \/>\nDraw the labelled block diagram of microcontroller.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57784\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-7.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 7\" width=\"533\" height=\"363\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-7.png 533w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-7-300x204.png 300w\" sizes=\"(max-width: 533px) 100vw, 533px\" \/><\/p>\n<p>Question 25.<br \/>\nHow do you represent.<br \/>\n(i) logical AND<br \/>\n(ii) logical OR<br \/>\n(iii) logical NOT operators in C programming?<br \/>\nAnswer:<br \/>\nInternet is worldwide interconnection of millions of computers by means of complex network which serves as an inexpensive communication medium.<\/p>\n<ol>\n<li>Long-distance telephone and cable TV systems.<\/li>\n<li>To provide LAN connections.<\/li>\n<li>Closed-circuit TV systems used for security.<\/li>\n<li>Aircraft communication and controls.<\/li>\n<\/ol>\n<p>Question 26.<br \/>\nWhat is internet? Write few applications of fibre optic communication system.<br \/>\nAnswer:<br \/>\nInternet<\/p>\n<ol>\n<li>Using internet, access to required information is easy, economical, reliable and fast.<\/li>\n<li>With E-mail, messages reach destination with in minutes.<\/li>\n<li>E-mail can be sent and received from any place.<\/li>\n<li>Using E-mail, message can be sent to several persons simultaneously.<\/li>\n<\/ol>\n<p>Applications<\/p>\n<ol>\n<li>Long-distance telephone and cable TV systems.<\/li>\n<li>To provide LAN connections.<\/li>\n<li>Closed-circuit TV systems used for security.<\/li>\n<li>Aircraft communication and controls.<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; D<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any Three questions:\u00a0 \u00a03 x 5 = 15<\/span><\/p>\n<p>Question 27.<br \/>\nFor the given CE amplifier circuit using silicon transistor. Calculate<br \/>\ni) Voltage across R<sub>2<\/sub><br \/>\nii) I<sub>E<br \/>\n<\/sub>iii) Z<sub>in(base)<\/sub><br \/>\niv) Z<sub>in <\/sub>(and)<br \/>\nv) Z<sub>o<br \/>\n<\/sub>Given \\(\\text { Given } r^{\\prime} e=\\frac{26 m V}{I_{E}}\\) and \u03b2 = 100<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57783\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-8.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 8\" width=\"394\" height=\"278\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-8.png 394w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-8-300x212.png 300w\" sizes=\"(max-width: 394px) 100vw, 394px\" \/><br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57782\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-9.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 9\" width=\"342\" height=\"217\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-9.png 342w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-9-300x190.png 300w\" sizes=\"(max-width: 342px) 100vw, 342px\" \/><br \/>\nZ<sub>in<\/sub> = R<sub>1<\/sub> || R<sub>2<\/sub>|| Z<sub>in(base)<\/sub> =1.38k\u03a9 = 2k\u03a9<br \/>\nZ<sub>b<\/sub> =Rc || R<sub>L<\/sub> =(10k\u03a9) || (10k\u03a9) = 5k\u03a9<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 28.<br \/>\nCalculate the output voltage if V<sub>1<\/sub>= 300mV and V<sub>2<\/sub> = 700 mV in the following circuit.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57781\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-10.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 10\" width=\"526\" height=\"241\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-10.png 526w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-10-300x137.png 300w\" sizes=\"(max-width: 526px) 100vw, 526px\" \/><br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57780\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-11.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 11\" width=\"551\" height=\"143\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-11.png 551w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-11-300x78.png 300w\" sizes=\"(max-width: 551px) 100vw, 551px\" \/><\/p>\n<p>Question 29.<br \/>\nA Hortley oscillator oscillates at 15 KHz. If the capacitor in the tank circuit has a value of 0.01 pF and one of the inductors value is 1 mH, calculate the value of the other inductor.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57779\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-12.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 12\" width=\"523\" height=\"140\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-12.png 523w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-12-300x80.png 300w\" sizes=\"(max-width: 523px) 100vw, 523px\" \/><br \/>\nL<sub>1<\/sub> + L<sub>2<\/sub> = L<sub>T<\/sub>= 11.25 mH<br \/>\nL,<sub>1<\/sub>+ 1mH= 11.25mH, L<sub>1<\/sub> = 10.25mH<\/p>\n<p>Question 30.<br \/>\nA frequency modulated signal is given by 10sin[6 x 108t + 5 sin 1250t].<br \/>\nDetermine<br \/>\ni) Carrier frequency<br \/>\nii) Modulating frequency<br \/>\niii) Modulation index<br \/>\niv) Maximum deviation<br \/>\nv) Carrier swing<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57778\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-13.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 13\" width=\"360\" height=\"181\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-13.png 360w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-13-300x151.png 300w\" sizes=\"(max-width: 360px) 100vw, 360px\" \/><br \/>\nMaximum frequency deviation = \u03b4= m<sub>f <\/sub>f<sub>m<\/sub>. = 5 x 199.04= 995.2 Hz<br \/>\nCarrier swing = 28 = 2(995.2) = 1990.4 Hz.<\/p>\n<p>Question 31.<br \/>\nSimplify the Boolean expression Y = sm(l,2,3,8,9,11,13) + sd(0,10,15) using K-map. Draw the logic diagram using NAND-gates to realise the simplified expression.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57777\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-14.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 14\" width=\"540\" height=\"194\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-14.png 540w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-14-300x108.png 300w\" sizes=\"(max-width: 540px) 100vw, 540px\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; E<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any Four questions.\u00a0 4 x 5 = 20<\/span><\/p>\n<p>Question 32.<br \/>\nGive the comparison between CB, CC and CE amplifiers on the performance characteristics.<br \/>\nAnswer:<\/p>\n<table border=\"2px\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"121\"><strong>Parameter<\/strong><\/td>\n<td style=\"text-align: center;\" width=\"129\"><strong>Types of Amplifiers.<\/strong><\/td>\n<td width=\"99\"><\/td>\n<td width=\"108\"><\/td>\n<\/tr>\n<tr>\n<td width=\"121\"><\/td>\n<td style=\"text-align: center;\" width=\"129\">CE<\/td>\n<td style=\"text-align: center;\" width=\"99\">CB<\/td>\n<td style=\"text-align: center;\" width=\"108\">CC<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Current gain<\/td>\n<td width=\"129\">High (P)<\/td>\n<td width=\"99\">Less than 1<\/td>\n<td width=\"108\">Highest (1+ p)<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Voltage gain<\/td>\n<td width=\"129\">High<\/td>\n<td width=\"99\">Moderate<\/td>\n<td width=\"108\">Low<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Power gain<\/td>\n<td width=\"129\">Highest<\/td>\n<td width=\"99\">Moderate<\/td>\n<td width=\"108\">Moderate<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Phase shift<\/td>\n<td width=\"129\">180\u00b0<\/td>\n<td width=\"99\">0\u00b0<\/td>\n<td width=\"108\">0-<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Input impedance<\/td>\n<td width=\"129\">Moderate<\/td>\n<td width=\"99\">Low<\/td>\n<td width=\"108\">High<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Output impedance<\/td>\n<td width=\"129\">Moderate<\/td>\n<td width=\"99\">High<\/td>\n<td width=\"108\">Low<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Bandwidth<\/td>\n<td width=\"129\">Narrow<\/td>\n<td width=\"99\">Wide<\/td>\n<td width=\"108\">Wide<\/td>\n<\/tr>\n<tr>\n<td width=\"121\">Applications<\/td>\n<td width=\"129\">Used as AF amplifies<\/td>\n<td width=\"99\">Used as RF amplifies<\/td>\n<td width=\"108\">impedance<br \/>\nmatching<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 33.<br \/>\nWhat is a logarithmic amplifier? With circuit diagram, obtain an expression for output voltage of logarithmic amplifier using OP-amp.<br \/>\nAnswer:<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57776\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-15.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 15\" width=\"426\" height=\"244\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-15.png 426w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-15-300x172.png 300w\" sizes=\"(max-width: 426px) 100vw, 426px\" \/><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57775\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-16.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 16\" width=\"381\" height=\"530\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-16.png 381w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-16-216x300.png 216w\" sizes=\"(max-width: 381px) 100vw, 381px\" \/><br \/>\n\u2234 The output voltage is proportional to natural logarithm of input voltage.<\/p>\n<p>Question 34.<br \/>\nWith a labelled block diagram, explain the working of AM transmitter.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57774\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-17.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 17\" width=\"584\" height=\"205\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-17.png 584w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-17-300x105.png 300w\" sizes=\"(max-width: 584px) 100vw, 584px\" \/><br \/>\nAnswer:<br \/>\nThe transmitter performs modulation and raises the power level of a modulated wave to required level for effective radiation.<\/p>\n<p>The microphone converts the audio signal into an equivalent audio electrical signal. The output is filtered to have a bandwidth of 10kHz.<\/p>\n<p>The pre amplifier provides the required voltage amplification. The driver amplifier increases the power level of the signal as required by the high power modulation amplifier.<\/p>\n<p>The high frequency carrier wave is generated by a crystal oscillator which has highest frequency stability. The Carrier signal is fed to a buffer amplifer which is a low gain, high input impedance amplifer which isolates crystal oscillator and the power level of the carrier signal to drive the modulated class C amplifier.<\/p>\n<p>The modulating audio signal and carrier signal are applied to the modulator. The collector modulation is used for modulation in high power transmitter. The transmitting antenna transmits the signal.<\/p>\n<p>Question 35.<br \/>\nExplain with circuit diagram and truth table, the working of clocked SR flip flop using NAND gates.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57773\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-18.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 18\" width=\"326\" height=\"106\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-18.png 326w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-18-300x98.png 300w\" sizes=\"(max-width: 326px) 100vw, 326px\" \/><br \/>\nIt is constructed using inverters inserted into inputs of cross coupled NAND gates.<br \/>\nWorking:<\/p>\n<ol>\n<li>When S=0, R=0, it does not respond and hence outputs Q and Q will remain in their previous state. This is called HOLD condition.<\/li>\n<li>When S=1 and R=0, the output Q and Q change to 1 and 0. This condition is called SET state.<\/li>\n<li>When S=0, R=1, the output Q and Q change to 0 and 1. This condition is called RESET state.<\/li>\n<li>When S=1 and R=l, it drives output Q and Q both to HIGH which is FORBIDDEN or INVALID condition.<\/li>\n<\/ol>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57770\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-19.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 19\" width=\"684\" height=\"193\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-19.png 684w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-19-300x85.png 300w\" sizes=\"(max-width: 684px) 100vw, 684px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 36.<br \/>\nWrite an assembly language instructions to move 45H into register A and 5 EH into register R<sub>0<\/sub>. Then add them together and save the result in R<sub>t<\/sub>. What is the content of R, after the execution of the program.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57769\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-20.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 20\" width=\"290\" height=\"281\" \/><\/p>\n<p>Question 37.<br \/>\nWrite a C program to accept three integer and find their sum and average. Find the sum and average of 5, 6 and 7.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-57768\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-21.png\" alt=\"2nd PUC Electronics Previous Year Question Paper March 2017 21\" width=\"285\" height=\"477\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-21.png 285w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/02\/2nd-PUC-Electronics-Previous-Year-Question-Paper-March-2017-21-179x300.png 179w\" sizes=\"(max-width: 285px) 100vw, 285px\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can Download 2nd PUC Electronics Previous Year Question Paper March 2017, Karnataka 2nd PUC Electronics Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Electronics Previous Year Question Paper March 2017 2nd PUC Electronics Previous Year Question Paper March 2017 &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/24813"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=24813"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/24813\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=24813"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=24813"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=24813"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}