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3 years ago

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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3 years ago

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

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3 years ago

SOMEONE PLEASE HELP!!

Arisa [49]

Hello!

$\large\boxed{\text{C. 7,350,000 J}}$

Use the equation:

$PE = mgh$

Where:

m = mass of the object (kg)

g = acceleration due to gravity (≈9.8 m/s)

h = height above ground (m)

Plug the given values into the equation:

PE = 7500 · 9.8 · 100

PE = 7,350,000 Joules.

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3 years ago

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Ksenya-84 [330]

False. That description fits the wave's 'frequency'.
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3 years ago

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Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separ

Roman55 [17]

Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

The Coulomb's force between two charges is given by;

$F = \frac{kq^2}{r^2} \\\\let \ kq^2 \ be \ constant\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\increasing \ the \ force \ between \ them \ by \ factor \ of \ 25\\\\(F_2 = 25F_1)\\\\r_2^2 = \frac{F_1r_1^2}{25F_1}\\\\r_2^2 = \frac{r_1^2}{25}\\\\r_2 = \sqrt{\frac{r_1^2}{25} }\\\\ r_2 = \frac{r_1}{5}$

r₂ = 0.2r₁

Therefore, the separation between the charges was decreased by a factor of 0.2.

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3 years ago