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2 years ago

2. A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.

Physics

1 answer:

svp [43]2 years ago

8 0

(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(d) The average power dissipated by the resistor is 38.4 W.

<h3>Maximum potential difference</h3>

Vrms = 0.7071V₀

where;

V₀ is peak voltage

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

<h3> rms current through the resistor </h3>

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

<h3>maximum current through the resistor </h3>

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

<h3> Average power dissipated by the resistor</h3>

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

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A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

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The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

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F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

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$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

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$F=qE$ is the magnitude of the electric force. Since

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$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

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$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

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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

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

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