SSC JE Electrical Engineering 2017 Set 4 Paper

Question : 140 of 200

Marks: +1, -0

In the range of an analog transducer is 0 to 10V. then for a resolution of 5mV. the bits of ADC will be_____

Solution:

Resolution =

{\Range \of \Input \votlage}/{2^n-1}

={V_ \max-V_\in}/{2^n-1}

5 ×10^{-3}={10- 0}/{2^n-1}

{2^{n-1}}=10/5×10^3

2^n=2000+1

n = 11

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