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prisoha [69]
3 years ago
14

The value read at an analog input pin using analogRead() is returned as a binary number between 0 and the maximum value that can

be stored in [X] bits. 1. This binary number is directly linearly proportional to the input voltage at the analog pin, with the smallest and largest numbers returned corresponding to the minimum and maximum ADC input values, respectively. At a Vcc of 3.3 V, analogRead(A0) returns a value of 1023. Approximately what voltage is present at the pin A0 on the MSP430F5529?
Engineering
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer: The approximate voltage would be the result from computing analogRead(A0)*3.3 V / 1023

Explanation:

Depending on the binary read of the function analogRead(A0) we would get a binary value between 0 to 1023, being 0 associated to 0V and 1023 to 3.3V, then we can use a three rule to get the X voltage corresponding to the binary readings as follows:

3.3 V ---------> 1023

X  V------------>analogRead(A0)

Then X=\frac{3.3 \times analogRead(A0)}{1023} Volts

Thus depending on the valule analogRead(A0) has in bits we get an approximate value of the voltage at pin A0, with a precission of 3,2mV approximately (3.3v/1023).

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A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
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Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

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