The power factor of the circuit is 0.7071
<h3>What is Power factor?</h3>
Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA). Apparent power, also known as demand, is the measure of the amount of power used to run machinery and equipment during a certain period. It is found by multiplying (kVA = V x A). The result is expressed as kVA units.
Power factor is expressed as the cosine of the phase angle.
The phase angle is given as 45°
Therefore the power factor in the circuit is cos45° = 0.7071
learn more about power factor from
brainly.com/question/25543272
#SPJ1
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:
0.9104
Explanation:
Suitable technology can tell you the probability.
P(-1.5≤Z≤2) ≈ 0.9104
__
A phone app gives the probability as 0.9104426667829628.