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Basile [38]
3 years ago
5

The rms current output in a circuit is 4.67 A. What is the maximum current?

Physics
2 answers:
jek_recluse [69]3 years ago
6 0

Answer:

I_O = \sqrt2 I_{rms}

I_O = \sqrt2 (4.67) = 6.6 A

Explanation:

As we know that maximum current in AC circuit is related to rms value of the current as per the equation below

I_{rms} = \frac{I_O}{\sqrt2}

here the value of maximum current or current amplitude is at numerator.

So in order to find maximum current or current amplitude we can say

I_O = \sqrt2 I_{rms}

now plug in the value of rms current in the above equation

I_O = \sqrt2(4.67)

I_O = 6.6 A

so maximum current is 6.6 A

Mila [183]3 years ago
5 0

Answer:Electrical power consumed by a resistance in an AC circuit is different to the power consumed by a reactance as reactances do not dissipate energy

   

In a DC circuit, the power consumed is simply the product of the DC voltage times the DC current, given in watts. However, for AC circuits with reactive components we have to calculate the consumed power differently.

Electrical power is the “rate” at which energy is being consumed in a circuit and as such all electrical and electronic components and devices have a limit to the amount of electrical power that they can safely handle. For example, a 1/4 watt resistor or a 20 watt amplifier.

Electrical power can be time-varying either as a DC quantity or as an AC quantity. The amount of power in a circuit at any instant of time is called the instantaneous power and is given by the well-known relationship of power equals volts times amps (P = V*I). So one watt (which is the rate of expending energy at one joule per second) will be equal to the volt-ampere product of one volt times one ampere.

Explanation:

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A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp
Sergeu [11.5K]

Answer:

U=1.29\times 10^{-7}\ J

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm  ,K₁= 4.7

d₂=2.1 mm  , K₂=2.6

We know that capacitance given as

C_1=\dfrac{K_1\varepsilon _oA}{d_1}

C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_1=1.26\times 10^{-10}\ F

C_2=\dfrac{K_2\varepsilon _oA}{d_2}

C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_2=0.701\times 10^{-10}\ F

Net capacitance

C=\dfrac{C_1C_2}{C_1+C_2}

C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F

C=4.5\times 10^{-11}\ F

We know that stored energy given as

U=\dfrac{CV^2}{2}

V= 76 V

U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J

U=1.29\times 10^{-7}\ J

3 0
3 years ago
At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

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c. Velocity of particle at origin

Using conversation of momentum

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We are told that M2 is initially at rest, then, V2=0

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We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

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Answer

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Explanation:

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