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Explanation:
Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.
This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.
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<u>Example</u>
Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...
mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)
mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs
mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0
You will note that the matrix of equation coefficients is symmetric.
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In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.
Answer:
work will be positive when it is under polytropic expansion process
Explanation:
It states a polytropic process with n equal to 1.67. there is a polytropic expansion that mean work is positive and if it was polytropic compression then it would be negative
Also work during the process of polytropic is given as
the work will be positive when it is under the polytropic expansion process
Answer:
a = 40
b = 29
Explanation:
Give a place holder for the numbers that we don't know.
Lets call the two numbers a and b.
From the given info, we can write an expression and solve it:
"one number is 11 more than another number"
a = 11 + b
from this, we know that a > b.
''three times the larger number exceeds four times the smaller number by 4"
3a = 4b + 4
Now we have 2 equations, we can use them to solve using whatever method you want.
a = 11 + b
3a = 4b + 4
I will be using matrices RREF to solve for this.
a - b = 11
3a - 4b = 4


a = 40
b = 29