Then number of bits are '4'
Resolution
F1 > F2 > F3
j = light intensity = constant
Photocurrent Vs Anode voltage with frequency and incident light as a parameter.
The light intensity is constant.
The area of the shaded region
.
Number of samples per period = 10p
This is not a rational number. Hence signal is non periodic.
? VP = 3.46 x 108 m/s.
n = 4
? 8R = 32
R = 4 K?.
i1 + ai1 = i2
V2 = i2R2
Putting a = - 1
= 0 V.
= 9.03 x 1028 electrons/m3
Since for tungsten the atomic and the molecular weights are the same. Therefore for tungsten
EF = 3.64 x 10-19(?)2/3
= 3.64 x 10- 19(9.03 x 1028)2/3
EF = 7.326 eV.
m = 2.5 for electrons and 2.7 for holes in Si.
Also, one input of the EXOR gate is tied to zero therefore 'f' will be directly transferred to the output.
The logic gates will introduce gate delays which will lead to the answer .
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