Match each context to the type of the law that is most suitable for it.
1 answer:
You might be interested in
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.
__
<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.
__
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:
1788.9 MPa
Explanation:
The magnitude of the maximum stress (σ) can be calculated usign the following equation:
<u>Where:</u>
<em>ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)</em>
<em>σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi) </em>
<em>2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in) </em>
Hence, the maximum stress (σ) is:
Therefore, the magnitude of the maximum stress is 1788.9 MPa.
I hope it helps you!