Design the necessary circuit using Logisim to implement the situation described above. Use Kmaps for simplification. Be VERY car
eful to get the correct functions for your four outputs before simplifying and designing the circuit with Logisim. You should minimize the entire circuit. This means you should look at each function AND its complement to see which combination will give you the minimal circuit. For example, using the functions that represent UL, UR (the complement of UR), LR, and LL (with NOT gates after UR and LL of course) might give the minimal circuit overall.Your circuit should have four inputs and four LED outputs. Your outputs should be such that the LEDs are arranged to light up the dots in their actual places (stack the 4 LEDs into a square.)All inputs should be labeled (in Logisim, not by hand). You can label outputs if you wish, but since you are placing the LEDs in the correct position, that isn't necessary. Please use the labeling for input (and output if you choose) as given above so everyone's circuit is labeled the same.You should also include your name as a label on the circuit.Test your circuit to be sure it is working correctly.Name the file using your first initial and last name followed by 4-2, for example lnull4-2.circ.
2 answers:
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Answer:
The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V
Explanation:
As the complete question is not given the complete question is found online and is attached herewith.
By applying KCL at node 1
Also
Now applying KVL on loop 1 as indicated in the attached figure
Similarly for loop 2
So the system of equations become
Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as
The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.