The Type K thermocouple has a sensitivity of about 41 micro-Volts/℃, i. e. for each degree difference in the junction temperature, the output changes by 41 micro-Volts. If you have a 32-bit ADC, what is the smallest temperature change you can detect if the ADC range is 10 V?

5.68*10^-5 °C

Explanation:

ADC is an acronym that means analog to digital conversion.

The numbers of bit in any ADC is the level of Voltage it possess.

The question states 32 bits, this means that it's total voltage levels is 2^32

Also, the question gives us a range of 10 V for the ADC. Now we can evaluate the resolution of the voltage as:

Δv = 10 / (2^32)

Δv = 10 / 4.29^9

Δv = 2.33*10^-9

The thermocouple sensitivity is 41 micro volts/°C

Thermocouple sensitivity = Δv/ΔT, where ΔT is the temperature change

ΔT = Δv/thermocouple sensitivity

ΔT = 2.33*10^-9 / 41*10^-6

ΔT = 5.68*10^-5 °C

Thus, the smallest temperature change is 5.68*10^-5 °C

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.