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:
[tex]Z_{R}[/tex] = R
R = resistance
[tex]Z_{L}[/tex] = jωL
ω = voltage source angular frequency, L = inductance
[tex]Z_{C}[/tex] = -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:
[tex]Z_{R}[/tex] = 217Ω
[tex]Z_{L}[/tex] = j(220)(0.875) = j192.5Ω
[tex]Z_{C}[/tex] = -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 = [tex]Z_{R}[/tex] + [tex]Z_{L}[/tex] + [tex]Z_{C}[/tex]
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:
[tex]v_{R}[/tex] = I×[tex]Z_{R}[/tex]
[tex]v_{L}[/tex] = I×[tex]Z_{L}[/tex]
[tex]v_{C}[/tex] = I×[tex]Z_{C}[/tex]
Given values:
I = (0.0569∠1.147+220t)A
[tex]Z_{R}[/tex] = 217Ω = (217∠0)Ω
[tex]Z_{L}[/tex] = j192.5Ω = (192.5∠π/2)Ω
[tex]Z_{C}[/tex] = -j673.4Ω = (673.4∠-π/2)Ω
Plug in and calculate each component's voltage:
[tex]v_{R}[/tex] = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V
[tex]v_{L}[/tex] = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V
[tex]v_{C}[/tex] = (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
[tex]v_{R}[/tex] = (12.35∠1.147+220t)V
[tex]v_{L}[/tex] = (10.95∠2.718+220t)V
[tex]v_{C}[/tex] = (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
[tex]v_{R}[/tex] = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V
[tex]v_{L}[/tex] = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V
[tex]v_{C}[/tex] = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V
The root mean square voltage (Vrms) of the circuit is 29.76 V.
The voltage in the inductor (VL) is 19.83 V.
The voltage in the capacitor (Vc) is 0.069V.
The voltage in the resistor (VR) is 22.35 V.
The rms voltage is calculated as follows;
V(rms) = -V₀sin(ωt)
V(rms) = -30 x sin(220 x 22 x 10⁻³)
V(rms) = 29.76 V
The overall opposition to the flow of current is calculated as follows;
Xl = ωL
Xl = 220 x 0.875 = 192.5 ohms
Xc = 1/ωC
Xc = (1) / (220 x 6.75 x 10⁻³)
Xc = 0.673 ohm
[tex]Z = \sqrt{R^2 + (X_l - X_c)^2} \\\\Z = \sqrt{217^2 + (192.5- 0.673)^2}\\\\Z = 289.63 \ ohm[/tex]
[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\I_{rms} = \frac{29.76}{289.63} \\\\I_{rms} = 0.103 \ A[/tex]
VL = IXl
VL = (0.103) x 192.5
VL = 19.83 V
Vc = IXc
Vc = 0.103 x 0.673
Vc = 0.069 V
VR = IR
VR = 0.103 x 217
VR = 22.35 V
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