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bagirrra123 [75]
3 years ago
11

A series RLC circuit with L = 13 mH, C = 3.6 µF, and R = 3.2 ohms is driven by a generator with a maximum emf of 120 V and a var

iable angular frequency, w ?. w= angular freq
(a) Find the resonant (angular) frequency ? w0.
________ rad/s

(b) Find I-rms at resonance.
________A

(c) Find the capacitive reactance XC in ohms.
_______ohms

Find the inductive reactance XL in ohms.
_______ohms

(d) Find the impedance Z. (Give your answer in ohms.)
_______ohms

Find I-rms.
_______A

(e) Find the phase angle (in degrees).
_______Degree
Physics
1 answer:
TEA [102]3 years ago
3 0

Answer: a) 4623 rad/sec b) 26.5 A c) 60 ohms. d) 3.2 ohms. e) 0º

Explanation:

a) At resonance, the circuit behaves like it were only resistive, so the reactive part of the impedance, is equal to 0.

So, the capacitive reactance and the inductive reactance are equal each other , as follows:

Xc = Xl  ⇒ ω₀L = 1/ω₀C  ⇒ ω₀ = 1/√LC

Replacing by the values of L and C, we get the value for ω₀, the angular frequency at resonance, as follows:

ω₀ = 4623 rad/sec

b) At resonance, the current can be obtained applying the Ohm´s Law as for any resistive-only circuit, as follows:

Irms = Vrms / R = (√2/2) Vmax / R = 85 V / 3.2 Ohms = 26.5 A

c) Xc = 1/ω₀C = 60 Ohms.

   Xl = ω₀L = 60 Ohms.

d) At resonance, as the total reactance becomes 0, the only component of the impedance that remains is the resistive part, so we can write the following : Z= R = 3.2 Ohms.

e) As the circuit behaves like if it were resistive only, voltage and current are in phase at any point of the circuit, so θ = 0º.

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While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
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Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

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3 years ago
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