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3 years ago

9

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

t which electrical energy is dissipated in the resistor is 216 W. (a) What is the impedance Z of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?

1 answer:

qaws [65]3 years ago

7 0

Answer:

impedance Z = 416.66 ohm

voltage across inducance V = 346.99 V

power factor = 0.720

Explanation:

given data

resistor R = 300

voltage amplitude = 500 V

resistor = 216 W

to find out

impedance and amplitude of the voltage and power factor

solution

we apply here average power formula that is

average power = I²×R ............1

I = $\frac{Vrms}{Z}$

so

average power = ( $\frac{Vrms}{Z}$ )²×R

Vrms = $\frac{1}{\sqrt{2} }$ × Vmax

Z = V × $\sqrt{\frac{R}{2P} }$

Z = 500 × $\sqrt{\frac{300}{2*216} }$

impedance Z = 416.66 ohm

and

we know voltage across inductor is here express as

V = I × X .............2

so here X will be by inductance

Z² = R² + (X)²

(X)² = 416.66² - 300²

X = 289.15 ohm

and I = $\frac{V}{Z}$

I = $\frac{500}{416.66}$

I = 1.20 A

so from equation 2

V = 1.20 × 289.15

voltage across inducance V = 346.99 V

and

average power = Vmax × Imax × cos∅

tan∅ = $\frac{289.15}{300}$

tan∅ = 43.95°

so power factor is

power factor = cos43.95°

power factor = 0.720

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Is it possible to have a charge of 5 x 10-20 C? Why?

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1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

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Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

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If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

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Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

6)

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- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

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$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

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7)

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Comparing the 2 forces,

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8 0

3 years ago

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Answer:

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Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

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429 / 1000 = 0.429 L of water

3 0

3 years ago