1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Inessa05 [86]
3 years ago
8

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

mplitude 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?
Physics
1 answer:
Nataly [62]3 years ago
4 0

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:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -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:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -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_{R} + Z_{L} + Z_{C}

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:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (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

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (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

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

You might be interested in
"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after
wel

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

3 0
3 years ago
An atom is.......Help?
Dmitry_Shevchenko [17]
The smallest level of ordination in living things. Every living thing is made up of atoms. Atoms are made up of protons, neutrons, and electrons.
5 0
2 years ago
Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. the tide measures 5.15 ft at
RSB [31]
<span>f(x) = 5.05*sin(x*pi/12) + 5.15

   First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

   So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
  f(x) = 5.05*sin(x*pi/12) + 5.15

   The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
 f(x) = 5.05*sin(x*pi/12 + C) + 5.15
  where
 C = Phase correction offset.

   But we don't need it for this problem, so the answer is:
 f(x) = 5.05*sin(x*pi/12) + 5.15

   Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
7 0
3 years ago
We can use energy principles to make ____ predictions.
stira [4]
Compatible and speedy
5 0
3 years ago
A box of mass 14 kg sits on an inclined surface with an angle of 52degrees. What is the component of the weight of the box along
kolezko [41]
First, we calculate for the weight of the object by multiplying the given mass by the acceleration due to gravity which is equal to 9.8 m/s²
                       Weight = (14 kg)(9.8 m/s²)
                          Weight = 137.2 N
The component of the weight that is along the surface of the inclined plane is equal to this weight times the sine of the given angle. 
                         Weight = (137.2 N)(sin 52°)
                               weight = 108.1 N
5 0
3 years ago
Read 2 more answers
Other questions:
  • A homemade capacitor is assembled by placing two 10-in. pie pans 2 cm apart by a piece of glass and connecting them to the oppos
    11·1 answer
  • Which of the following is an example of an object that could have a net force greater than zero acting on it?
    10·2 answers
  • What are the 4 rules of seismic waves?
    5·1 answer
  • Which is not an example of how an object gains elastic potential energy by stretching?
    11·1 answer
  • A car travels 60 miles in the first 2 hours and 68 miles in the next 2 hours
    5·1 answer
  • Which action will cause the induced current to decrease or remain constant?
    6·1 answer
  • The supersonic airliner Concorde is 62.1 m long when sittingon
    13·1 answer
  • The front and rear sprockets on a bicycle have radii of 8.40 and 4.91 cm, respectively. The angular speed of the front sprocket
    10·2 answers
  • PLS ANSWER DUE LATER TODAY!!!
    9·1 answer
  • 8. Where are the ribosomes usually located in plant and animal cells?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!