A217 ω resistor, a 0.875 h inductor, and a 6.75 μf capacitor are connected in series across a voltage source that has voltage amplitude 30.0 v and an angular frequency of 220 rad/s . what is v at t= 22.0 ms ? what is vr at t= 22.0 ms ? what is vl at t= 22.0 ms ? what is vc at t= 22.0 ms ? what is vr? what is vl? what is vc?

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

spectral lines lyman is one set

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s