As per Bernuolli's Theorem total energy per unit mass is given as
now from above equation
now by above equation
Part B)
Now energy per unit weight
What is the electric flux φ through each of the six faces of the cube? Use ϵ0 for the permittivity of free space?
1 answer:
If total charge Q is enclosed in the surface of cube
then we will have say that total flux linked with all surfaces of the cube will be given by
since the position of charge is symmetric with respect to the center of the cube so here this whole flux will be equally linked with each of the face of the cube
So here total 6 faces of the cube is there and the total flux of the charge will equally divide in these 6 faces
so flux linked with each face is given by formula
so each face will have above flux due to central position of charge Q
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For an AC circuit:
I = V/Z
V = AC source voltage, I = total AC current, Z = total impedance
Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.
For a resistor R, inductor L, and capacitor C, their impedances are given by:
= R
R = resistance
= jωL
ω = voltage source angular frequency, L = inductance
= -j/(ωC)
ω = voltage source angular frequency, C = capacitance
Given values:
R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s
Plug in and calculate the impedances:
= 217Ω
= j(220)(0.875) = j192.5Ω
= -j/(220×6.75×10⁻⁶) = -j673.4Ω
Add up the impedances to get the total impedance Z, then convert Z to polar form:
Z = +
+
Z = 217 + j192.5 - j673.4
Z = (217-j480.9)Ω
Z = (527.6∠-1.147)Ω
Back to I = V/Z
Given values:
V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)
Z = (527.6∠-1.147)Ω (from previous computation)
Plug in and solve for I:
I = (30.0∠0+220t)/(527.6∠-1.147)
I = (0.0569∠1.147+220t)A
To get the voltages of each individual component, we'll just multiply I and each of their impedances:
= I×
= I×
= I×
Given values:
I = (0.0569∠1.147+220t)A
= 217Ω = (217∠0)Ω
= j192.5Ω = (192.5∠π/2)Ω
= -j673.4Ω = (673.4∠-π/2)Ω
Plug in and calculate each component's voltage:
= (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V
= (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V
= (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V
Now we have the total and individual voltages as functions of time:
V = (30.0∠0+220t)V
= (12.35∠1.147+220t)V
= (10.95∠2.718+220t)V
= (38.32∠-0.4238+220t)V
Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:
V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V
= 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V
= 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V
= 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V