Correct Answer - Option 1 : 0.317 V and 12.06 V

__Concept:__

**Resolution **

The **ADC resolution** is defined as the smallest incremental voltage that can be recognized and thus causes a change in the digital output.

**Resolution = Step size**

__NOTE:__

If only LSB is present in the input sequence applied then the output voltage is defined as the step size.

V_{0} = Step Size.

__Calculation:__

Given supply voltage is 20 V and 6 bit ADC

Output voltage when only LSB is present is Resolution itself.

\({V_0} = \frac{{20}}{{{2^6} - 1}}\)

\({V_0} = \frac{{20}}{{63}}\)

**NOTE: **In the ESE prelims calculator is not allowed so we will do the approximate calculation.

Calculate 20/64 we get the result as

\(\frac{{20}}{{64}} = \frac{5}{{16}} = 0.3125\)

But denominator less than assumed so, the fraction value increase.

According to options it will be 0.317 V

**Converting into Decimal**

(100110)_{2} = ()_{10}

= 1 × 2^{5} + 0 × 2^{4} + 0 × 2^{3 }+ 1 × 2^{2 }+ 1 × 2^{1} + 0 × 2^{0}

= 32 + 4 + 2

= (38)_{10}

Now the output voltage is

V_{0} = 0.317 × 40

= 12.68 V

But the value is < 40 so answer also < 12.68 V

According to **options,** the answer is **12.06 V**