Problem 4 You are designing a circuit to drive LED1, using the following circuit. The datasheet for the LED specifies that VF =
2 V. Treat LEDs as diodes with the constant voltage model. (a) Find the expected current through the LED I(LED) when VF = 2 V (b) After testing, it is determined that some LEDs within a batch do not light up correctly in the circuit for part a. It is suspected that this is because there is some variation in among the LEDs in the batch. Find the maximum value of VF such that the circuit above stops working I(LED) drops to zero so you can characterize this problem and report it to the manufacturer.
1 answer:
Answer:
a) I_LED= 1/6 A b) Vf= 2.5V
Explanation:
Consider circuit in the attachment.
a) We will simplify current source in paraller with resistor to a voltage source in series with a resistor(see attachment 2)
Solving the circuit in attachment 2 using mesh analysis
-9+2I1+4(I1-I2)-4+2I1=0
8I1 - 4I2= 13 ............... eq 1
4+4(I2-I1)+ I2 + 2=0
4I1- 5I2 = 6 ............ eq 2
I1= 41/24 ; I2 = 1/6; I2= I_LED
b) Solving the circuit in attachment 2 again, this time I2=0
8I1 - 4I2= 13
8I1- 4(0)=13
I1= 13/8
Vf= 4(I1- I2) -4
I2=I_LED=0
Vf= 2.5 V
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Answer:
The answers are as follow:
a) 10 mm
b) 12.730 N/
c) 127.307 N/
d) 0.25
Explanation:
d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9
Force = F =1000 N
let us find initial area first, A1 = pi* = 78.55
using reduction in area formula : 0.9 = (A1 - A2 ) / A1
solving it will give, A2 = 0.1 A1 = 7.855
a) The specimen elongation is final length - initial length
50 - 40 = 10 mm
b) Engineering stress uses the original area for all stress calculations,
Engineering stress = force / original area = F / A1 = 1000 / 78.55
Engineering stress = 12.730
c) True stress uses instantaneous area during stress calculations,
True fracture stress = force / final area = F / A2 = 1000 / 7.855
True Fracture stress = 127.30
e) strain = change in length / original length
strain = 10 / 40 = 0.25
Answer:
The pH of the solution will be equal to 5.46
Explanation:
The dissociation reaction of HCl is equal to:
HCl → H+ + Cl-
To solve the exercise we must first convert the µmoles to moles using the following conversion factor:
3.5µmoles x = 3.5x
moles
Assuming a liter of solution, we can calculate the molar concentration by:
M =
Replacing:
M = = 3.5x
moles/L
As this acid dissociates completely, the concentration of protons and chloride will be equal to 3.5xmoles/L
The pH will be equal to:
pH = -log[H+]
Replacing:
pH = -log[3.5x] = 5.46