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Tom [10]
1 year ago
11

What boolean operation does the circuit compute?

Physics
1 answer:
vagabundo [1.1K]1 year ago
8 0

A circuit computes various types of Boolean operations. As in the question no specific circuit has been shown,so we have to know the fundamental logic operations.

Out of the various types of logic operations in Boolean algebra,the simple operations are- addition,subtractions,multiplications,division through logic gates.

Various logic or Boolean operations are  NOT GATE,OR GATE ,AND GATE .

There are other operations also like NAND GATE,NOR GATE and XOR GATE etc

NOT gate is the simple logic gate which inverts one quantity.it makes 1[high] to 0[low] and vice versa.

OR gate gives 1 [high] when one of the input is 1 and low when all the inputs are low.

AND operation gives 1 when all the inputs are high and low when one of the input is low

NAND operation is AND gate followed by a NOT gate

NOR operation is the OR gate followed by a NOT gate.

Remembering these simple rules we can get the idea of the operations involved in the circuit. Each circuit has different symbols and gives different results.

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7 0
2 years ago
A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
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solution:

consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm

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