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
R = resistance
 = jωL
 = jωL
ω = voltage source angular frequency, L = inductance
 = -j/(ωC)
 = -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
 = 217Ω
 = j(220)(0.875) = j192.5Ω
 = j(220)(0.875) = j192.5Ω
 = -j/(220×6.75×10⁻⁶) = -j673.4Ω
 = -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×
 = I×
 = I×
 = I×
Given values:
I = (0.0569∠1.147+220t)A
 = 217Ω = (217∠0)Ω
 = 217Ω = (217∠0)Ω
 = j192.5Ω = (192.5∠π/2)Ω
 = j192.5Ω = (192.5∠π/2)Ω
 = -j673.4Ω = (673.4∠-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)(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)(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
 = (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
 = (12.35∠1.147+220t)V
 = (10.95∠2.718+220t)V
 = (10.95∠2.718+220t)V
 = (38.32∠-0.4238+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
 = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V
 = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V
 = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V
 = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V
 = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V