Which two terrestrial planets have atmospheres with similar compositions?
2 answers:
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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
The beat frequency is the concept required to develop this process. The phenomenon is generated when you have two waves but their frequencies do not differ greatly. So the beat frequency will be the difference between those two waves. For this case we have the difference and one of the waves, therefore,
Our values are given as,
Using the formula of the frequency I will have,
Replacing we have,
The possible values are two: 1045Hz and 1055Hz