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A: A capacitor looks like a capacitor, inductor, and resistor in series at high frequency. Less
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A short
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Its an short circuit at very high frequencies. Look at the impedance for a capacitor and you'll notice that the frequency is inversely related to it. (also note that capacitor impedance is purely reactive). Therefore, as frequency increases the impedance approaches 0, which acts like a short. (an inductor is the opposite) Less
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Zero..since dc can't produce emf in windings.
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Zero. Because, Transformer never allow DC power.
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Core Gets Saturated. Lol !
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question was tricky, since the output was already getting registered. Panelist wanted to know wat should be done before its registered. I had no idea what to answer. Less
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Since output of one domain is already registered and we are sending signal to another domain means we have to reenergise the signal for further process as it is in our control. Please correct me if i was wrong Less
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Perform buffering or isolation of the signal if required as per another domain.
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start with middle move to 2nd and check twice and move to 3rd and then 4th and check 4th twice.. right? OR you start anywhere just move sequentially and check 2nd and 4th place twice Less
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middle to 2nd not needed
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Keep checking middle one, worst case delay is 4 check, and you should find it by then!! Less
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The microcontroller gets the address of the ISR from the interrupt vector table and jumps to it. It starts to execute the interrupt service subroutine, which is RETI (return from interrupt). ... First, it gets the program counter (PC) address from the stack by popping the top bytes of the stack into the PC. Less
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This was 3 years back, now the Projects has been taken to the next level, prepare yourself for more demanding Q&A Less
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The tank is never completely clean. The cleanliness is like a decaying exponential function which never quite reaches 0% contaminants. Less
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The last answer is true, but it is also useless in terms of an engineering project. A better answer to give is: do a few iterations of the decaying exponential to demonstrate your grasp on the problem. Then assign an good enough definition of completely clean and plug in numbers until you get around that number. Less
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Use adders connected in series adding in 1 for each bit in the word which is '1', and 0 for each bit which is '0'. Less
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lets assume the number is 8 bits . So use 8 2-to-1 muxes ( inputs as 0 and 1 ) and feed the connect the select switches to each bit of the number . So if the bit is high , 1 is propagated or else 0 is propagated .Check for the number of high mux outputs. Less
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There is a mathematical solution but in the interest of time, you could suggest tossing the toothpick a number of times and counting the number of crosses divided by the total number of tosses. Less
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The above solution is probably not ideal given that you did not demonstrate any thinking skills behind it. First you need to ask how large is the floor. If the floor is as large as the surface of the earth, then the probability is almost 0. Then the interviewer might define the floor as the space demarcated by the two lines. In this case, you need to define the position of the toothpick by its center position and its angle/orientation. The center of the toothpick as and its orientation follow a uniform distribution. The more difficult part is to analyze what combination of the two parameters counts as a 'crossing' event. For example, if the center of the toothpick is the middle between the two lines, then whatever the angle you have, the toothpick does not cross the two lines. Conversely, if the center of the toothpick falls on one of the lines, then it's going to cross the line no matter what. For the intermediate cases, use trigo to find an expression for the angle. In the end, you need to do an integral and get the probability out. Less
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Very good
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Very good