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Kobotan [32]
3 years ago
14

A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate

at which electrical energy is dissipated in the resistor is 286 W. What is the impedance Z of the circuit?
Physics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

Explanation:

impedance z=(XL^2+R^2)^1/2

power across te resistor ==i^2r

286/300

I=.976

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attribute is defined as a quality or characteristic of a person, place, or thing

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3 years ago
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Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,
Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41  ,  RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V ,  Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc =  364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -)  √ 150.74² 77.0²

RL = 473 (+ / -)  (129.58)

RL₁ = 34.41  ,  RL₂ = 602.58

d.

The higher value have the less angular frequency  

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0
3 years ago
How to make a paper airplane curve down and fly upside down?
Ainat [17]
Throw it sideways and try to make it spin around but it needs to be thrown high up then it should kinda glide down
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What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic th
asambeis [7]

Answer:

\lambda = 2.57 \times 10^{-11} m

Explanation:

Average velocity of oxygen molecule at given temperature is

v_{rms} = \sqrt{\frac{3RT}{M}}

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}

v_{rms} = 483.4 m/s

now for de Broglie wavelength we know that

\lambda = \frac{h}{mv}

\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}

\lambda = 2.57 \times 10^{-11} m

7 0
3 years ago
g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor
Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0
2 years ago
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