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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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6 0
3 years ago
A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
levacccp [35]

Answer:

Q = 8,345 * v

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

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8 0
4 years ago
A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a
uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

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Where, m_{h}= mass of halfback

m_{c}=mass of corner back

v_{h}= velocity of halfback

v_{c}= velocity of corner back

Put the value into the formula

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v=\dfrac{98\times4.2+85\times5.5}{98+85}

v=4.80\ m/s

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0
3 years ago
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