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iris [78.8K]
3 years ago
10

A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. What does th

e time constant for the circuit represent? A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. What does the time constant for the circuit represent? The time constant represents the time required for the current to reach 25% of the maximum current. The time constant represents the time required for the current to reach 63% of the maximum current. The time constant represents the time required for the current to reach 99% of the maximum current. The time constant represents the time required for the current to reach 37% of the maximum current. The time constant represents the time required for the current to reach 75% of the maximum current. The time constant represents the time required for the current to reach 50% of the maximum current.
Physics
1 answer:
Irina-Kira [14]3 years ago
4 0

Answer:

The time constant τ = L/R represent the time requiered for the curent to get value of 63 % of its maximun value

Explanation:

In a circuit RL ( serie circuit with a resistor R and the inductor L ) when a voltage is applied the inductor will have a reaction, such reaction consist in the creation of an electromotive force which will prevent the current to get its maximun value. The time constant which in the case of an inductor is

τ = L/R ,  represent the time requiered for the crrent to be 63 % of its final value

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Tres litros de oxigeno gaseoso a 15 grados centígrados y a presión atmosférica (1atm), se lleva a una presión de 10mm de Hg. ¿ c
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Answer:

Three liters of gaseous oxygen at 15 degrees Celsius and at atmospheric pressure (1atm), is brought to a pressure of 10mm Hg. What will be the volume of the gas now if the temperature has not changed?

Explanation:

Given that, the temperature is constant

Then, using gay Lussac law

P1•V1 / T1 = P2•V2 / T2

Since temperature is constant

Then, T1 = T2 = T and they cancels out

So, we are left with

P1•V1 = P2•V2

Given that, .

Initial volume

V1 = 3 litres.

Initial pressure

P1 = 1atm = 101325 Pa

Final pressure

P2 = 10mmHg = 1333.22 Pa

Then, we want to find the final volume V2

Make V2 subject of formula.

V2 = V1•P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litres

So, the final volume is 288 litres.

In Spanish

Dado que la temperatura es constante

Luego, usando la ley gay de Lussac

P1 • V1 / T1 = P2 • V2 / T2

Como la temperatura es constante

Entonces, T1 = T2 = T y se cancelan

Entonces, nos quedamos con

P1 • V1 = P2 • V2

Dado que, .

Volumen inicial

V1 = 3 litros.

Presión inicial

P1 = 1atm = 101325 Pa

Presión final

P2 = 10 mmHg = 1333.22 Pa

Entonces, queremos encontrar el volumen final V2

Hacer V2 sujeto de fórmula.

V2 = V1 • P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litros

Entonces, el volumen final es de 288 litros

8 0
4 years ago
a transmission-line cable, of length 3 km, consists of 19 strands of identical copper conductors, each 1.5 mm in diameter. becau
Ivan

the resistance of the cable is 582.9 ohms

we are given the length of the cable which is  3 km,  of  1.5 mm in, the diameter  and resistivity of copper which is 1.72 m

The formula  we are referring to for calculating the resistance of the  cable is

R = ρl/A.

As there are 19 strands of copper conductors, so the resistance will be

R = 19( ρl/A)

Here  ρ is the resisitivity =  1.72 , l is the length  = 3(1+0.05)*10³3= 3150 m

A=pie/4(1.5 x 10⁻³)^2 =1.766 x 10⁻⁶ =1.766 x 10^-6

Substituting the values in the formula  we  get

R = 19 ( 1.72*3150 )/1.766 x 10⁻⁶

 = 582.9 ohm

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1 year ago
In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn
kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

s=ut+\dfrac{1}{2}at_{A}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

t_{A}=\sqrt{\dfrac{30\times2}{1.6}}

t_{A}=6.12\ sec

We need to calculate the time for B

Using equation of motion

s=ut+\dfrac{1}{2}at_{B}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

t_{B}=5.48\ sec

We need to calculate the time for longest nap

Using formula for difference of time

t'=t_{A}-t_{B}

t'=6.12-5.48

t'=0.64\ s

Hence, B can take 0.64 sec for the longest nap .

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