1. The minimum number of equations required to analyze the circuit shown in the figure is





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MCQ-> In each of the following questions two rows of number are given. The resultant number in each row is to be worked out separately based on the following rules and the question below the row is to be answered. The operations of number progress from the left to right. Rules: (i) If an even number is followed by another even number they are to be added. (ii) If an even number is followed by a prime number, they are to be multiplied. (iii) If an odd number is followed by an even number, even number is to be subtracted from the odd number. (iv) If an odd number is followed by another odd number the first number is to be added to the square of the second number. (v) If an even number is followed by a composite odd number, the even number is to be divided by odd number.I. 84 21 13 II. 15 11 44 What is half of the sum of the resultants of the two rows ?....
MCQ->In this question, a column and a statement followed. Read the paragraph carefully and decide on the basis of that column.Moscow officials said that a pair of Russian astronauts cut out the material samples around a spy hole in a Soyuz spacecraft stationed at the International Space Station on Tuesday using knives and chin knives. It was discovered in August, after the last trip of the spacecraft, that the 2-meter pipe airbag on the Soyuz spacecraft was parked at international space. The Rascassasmos Space Agency said that the small but dangerous hole had been found on Earth or in the outer space was to cut the enclosed glue material, to analyze the model of the material and to pull it into a new tapper in that area.Senior astronauts fought and ultimately succeeded in their pursuit. Before this astronaut walk, the spacecraft could only examine the hole within the spacecraft. Unlike the International Space Station, the Soyce shuttle was not designed to repair and hold the spacecraft in the outer space, but did not have any of the handles to hold. The claim was that astronauts had cut the glue around the hole, but failed to collect the sample to analyze Arkal.Select the correct answer from the following A - The claim is entirely correct B - the claim may be correct C - can not determine the claim D - The claim is completely false.....
MCQ-> In each of the following questions two rows of numbers are given. The resultant number in each row is to be worked out separately based on the following rules and the questions below the rows of numbers are to be answered. The operations of numbers progress from left to right. Rules: (a) If an odd number is followed by a two digit even number then they are to be added. (b) If an odd number is followed by a two digit odd number then the second number is to be subtracted from the first number. (c) If an even number is followed by a number which is a perfect square of a number then the second number is to be divided by the first number. (d) If an even number is followed by a two digit even number then he first number is to be multiplied by the second number.15 11 20 400 8 12 10 If the resultant of the second set of a numbers is divided by the resultant of the first set of numbers what will be the outcome ?....
MCQ-> Mathematicians are assigned a number called Erdos number (named after the famous mathematician, Paul Erdos). Only Paul Erdos himself has an Erdos number of zero. Any mathematician who has written a research paper with Erdos has an Erdos number of 1.For other mathematicians, the calculation of his/her Erdos number is illustrated below:Suppose that a mathematician X has co-authored papers with several other mathematicians. 'From among them, mathematician Y has the smallest Erdos number. Let the Erdos number of Y be y. Then X has an Erdos number of y+1. Hence any mathematician with no co-authorship chain connected to Erdos has an Erdos number of infinity. :In a seven day long mini-conference organized in memory of Paul Erdos, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdos number. Nobody had an Erdos number less than that of F.On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdos number of the group of eight mathematicians to 3. The Erdos numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdos number of the group of eight to as low as 3.• At the end of the third day, five members of this group had identical Erdos numbers while the other three had Erdos numbers distinct from each other.• On the fifth day, E co-authored a paper with F which reduced the group's average Erdos number by 0.5. The Erdos numbers of the remaining six were unchanged with the writing of this paper.• No other paper was written during the conference.The person having the largest Erdos number at the end of the conference must have had Erdos number (at that time):
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