A un Senior Hardware Engineer le preguntaron...1 de agosto de 2014

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@Don you are correct up until your conclusion that "Each bulb will be a little dimmer than in parallel, but 100W still brighter." The 100W bulb is 10x dimmer if you assume resistance does not depend on current. Reality is even worse since R increases with current. Since the 10W bulb has 10x higher resistance than 100W bulb, it will absorb 10x as much power in the voltage divider: 100W bulb = 1A at 100V, or 100 ohms 10W bulb = 0.1A at 100V, or 1000 ohms 100W bulb + 10W bulb in series driven off 110V line is 1100 ohms carrying 0.1A of current. Power delivered to 10W bulb is P = I^2 x R = (0.1A)^2 x (1000 ohm) = 10W. Power delivered to 100W bulb is P = (0.1A)^2 x (100 ohm) = 1W. So the 100W bulb only receives 1W delivered power because the 10W bulb absorbs all the power in the divider. The 10W bulb will be brighter. Menos

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1. When connected in series: In a series connection, current flowing across each element is same. So when 40W bulb and 60W bulb are connected in series, same current will flow through them. To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation P=(I*I) R since current is same we can say that power dissipation will be higher for the bulb with higher resistance i.e. 40W bulb. Hence 40W bulb will glow brighter in series connection. 2. When connected in parallel: In a parallel connection, voltage across each element is same. So when 40W bulb and 60W bulb are connected in parallel, voltage across them will be same (100 V in the given case). To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation P=(V*V)/R since voltage is same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 60W bulb. Hence 60W bulb will glow brighter in parallel connection. Menos

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I have attended the same question, 40W & 100W bulb with 110V supply , all are them connected in serial Answer : Who is having more resistance, he will have more power drop also he is the brighter one. Here 100W will have less resistance and 40W bulb has more resistance, So 40W will be glow brighter than 100W bulb R1 = 110 x 110 / 100 = 2.2 ohm R2 = 110 x 110 / 40 = 302.5 ohm Answer is 40W bulb will be brighter than 100W bulb Menos

A un Senior Hardware Verification Engineer le preguntaron...30 de enero de 2010

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Write the number in binary and count the number of ones in that.If the number os ones is only 1 then it the number is indeed a power of 2 Menos

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For an integer n: If n is less than 1, return false. If the bitwise & of n and n-1 is 0, return true. Otherwise, return false. Menos

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I think the main idea is to use recursion function, for the integer which is larger than 0, if it is 1 return true, else return function(n-1) Menos

A un Senior Hardware Verification Engineer le preguntaron...30 de enero de 2010

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out = (A NAND (B NAND B))

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OUT = (A NAND (B NAND 1)) or out = (A NAND (B NAND B)) like what anonymous said. Menos

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out = ((A NAND A) NAND (A NAND A)) NAND (B NAND B)

A un Senior Hardware Verification Engineer le preguntaron...30 de enero de 2010

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Take 1 candy from bowl 1, 2 candies from bowl 2, 3 candies from bowl 3, and 4 candies from bowl; and weight them on the digital scale. If the total weight is 100 grams, the 9 gram candies are in bowl 5. If the total weight is 99 grams, the 9 gram candies are in bowl 1. If the total weight is 98 grams, the 9 gram candies are in bowl 2. If the total weight is 97 grams, the 9 gram candies are in bowl 3. If the total weight is 96 grams, the 9 gram candies are in bowl 4. Menos

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if weighing machine can weigh the bowls: Put 2 bowls on each side, if they weigh equal then last bowl has candies with 9 grams. If the bowl weigh different then weigh the 2 bowls with lower weight and find the lowest weight bowl. Menos

A un Senior Hardware Engineer le preguntaron...6 de marzo de 2019

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Oscilloscope. (Maybe I should be interviewing you?)

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Oscope and Spectrum Analyzer :)

A un Senior Hardware Engineer le preguntaron...5 de marzo de 2012

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Much easier than that: inline int isLE() { int v = 1; //stored in host order char *p = (char *)&v; //can access addresses &p[0], ..., &p[3] //LS byte stored at lower address in memory => LE return p[0]; } I'm sure there's a more efficient way, but that's the best I know. Menos

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To answer this question, you need to know about the union in C. 1) declare union { unsigned long ul; unsigned char tab[4] } un; 2) then: un.ul=0x12345678; for(i=0; i<4; i++) printf("%x ",un.tab[i]); Menos

A un Senior Hardware Engineer le preguntaron...4 de octubre de 2019

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Can you tell more about hour long coding interview? More detail on coding problem would help. Thank you! Menos

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You need to write an FSM. For whatever they ask!

A un Senior Hardware Engineer le preguntaron...12 de diciembre de 2009

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about 22 million in the main cabin depending on the installed options.

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The exact same number as the number of angels on the head of a pin. Go figure.

A un Senior Hardware Engineer le preguntaron...28 de octubre de 2018

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Functionally split the team members based on their background and strengths.

A un Senior Hardware Engineer le preguntaron...26 de octubre de 2017