Milton has been tracking the migrating patterns of whales in the northwest Atlantic Ocean for five years. He knows where and whe
2 answers:
You might be interested in
Answer:
- branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
- branch 2: i = 21.466∠63.435°; pf = 0.447 leading
- total: i = 31.693∠10.392° leading; pf = 0.984 leading
Explanation:
To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.
The inductive reactance is ...
The capacitive reactance is ...
<u>Branch 1</u>
The impedance of branch 1 is ...
Z1 = 8 +j4.99984 Ω
so the current is ...
I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°
The power factor is cos(-32.005°) ≈ 0.848 (lagging)
<u>Branch 2</u>
The impedance of branch 2 is ...
Z2 = 5 -j10 Ω
so the current is ...
I2 = 240/(5 +j10) ≈ 21.466∠63.435°
The power factor is cos(63.436°) ≈ 0.447 (leading)
<u>Total current</u>
The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.
It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)
It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°
The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)
__
The phasor diagram of the currents is attached.
_____
<em>Additional comment</em>
Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.
Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
Q = 5.06 x 10⁻⁸ m³/s
Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Answer:
Detailed solution is given below:
