Answer:
Explanation:
1) Resolution and uncertainty of both ADC ranges
a) Resolution (for +/- 5 mV range) = 5 mV / 2n where n = number of bits of ADC
Resolution = 5 mV / 28 = 5 mV / 256 = 0.0195 mV ..................1
uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*0.0195 mV = 0.00975 mV ..........2
b) Resolution (for +/- 5 V range) = 5 V / 2n where n = number of bits of ADC
Resolution = 5 mV / 28 = 5 V / 256 = 0.0195 V = 19.5 mV ..................3
uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*19.5 mV = 9.75 mV ..........4
2)Thermocouple sensitivity = ( Maximum output voltage - Minimum Output voltage) / (Maximum Temperature - Minimum Temperature)
Thermocouple sensitivity = (3.649mv - 0 ) / (70 - 0) = 0.0521 mV / Deg.C ............5
This is the required Thermocouple sensitivity
3) Water bath temperature is given as 57 deg.C
Hence voltage read by Thermocouple = Sensitivity*57 = 0.0521*57 mV = 2.9697 mV ........6
4)We need to use ADC with a range of +/- 5 mV range as ADC with +/- 5 V range can not do measurement as it's resolution is higher than output voltage.
ADC will measure voltage as 2.9695 mV ......................7
Assuming V1 is the anode and v2 the cathode (Anode is P region and Cathode is N)
Answer:
a) Reverse bias
b) Forward bias
c) Forward bias
Explanation:
Forward bias: It happens whenever the N region of the diode is more positive than the P region. Hence, the depletion zone increase ceasing the current through the circuit -> V1 -V2 < 0
Reverse bias: It happens whenever the P region of the diode is more positive than the N region. In this case, the depletion zone begins to shrink, if enough voltage is applied current could go through the circuit -> V1 - V2 > 0
a) V = V1 - V2 = 0 - 2 = -2 -> -2 is smaller than zero therefore, we have reverse bias
b) V = V1 - V2 = 4.5 - 2.8 = 1.7 -> 1.7 is greater than zero therefore, we have forward bias
c9 V = V1 - V2 = -1 - -1.3 = 0.3 -> 0.3 is greater than zero therefore, we have forward bias
Answer:
C. It is a continually growing field, and individuals trained in it should not have a problem finding employment.
Explanation:
D) O Water - Based Is the 1000000000000000000% correct answer
Answer:
10.203 Volts
Explanation:
For this problem, we need to understand that a series resistive circuit is simply a circuit with some type of voltage source and some resistors, in this case, R1 and R2.
First, we need to find the voltage in the circuit. To do this, we need to find the total resistance of the circuit. When two resistors are in series, you sum the resistance. So we can say the following:
R_Total = R1 + R2
R_Total = 570 Ω + 560 Ω
R_Total = 1130 Ω
Now that we have R_Total for the circuit, we can find the voltage of the circuit by using Ohm's law, V = IR.
V_Total = I_Total * R_Total
V_Total = 17.9 mA * 1130 Ω
V_Total = 20.227 V
Now that we have V_Total, we can find the voltage drop across each resistor by using Ohm's law once more. Note, that since our circuit is series, both resistors will have the same current (I.e., I_Total = I_1 = I_2).
V_Total = V_1 + V_2
V_Total = V_1 + I_2*R2
V_Total - I_2*R2 = V_1
20.227 V - (17.9 mA * 560 Ω) = V_1
20.227 V - (10.024 V) = V_1
10.203 V = V_1
Hence, the voltage drop across R1 is 10.203 Volts.
Cheers.
