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3 years ago

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DL: Activity 2.3 Resistance Complete the questions based on the Resistance presentation. a. All resistors _____ or _

______ the flow of electrons. b. As resistance __________________ current _______________________

1 answer:

Jlenok [28]3 years ago

5 0

<h2>Question:</h2>

DL: Activity 2.3 Resistance

Complete the questions based on the Resistance presentation.

a. All resistors ___________ or _____________ the flow of electrons.

b. As resistance __________________ current _______________________

<h2></h2><h2>Answer:</h2>

a) limits or opposes

b) increases, decreases

<h2>Explanation:</h2>

Resistors are electrical devices used to resist, limit, oppose or hinder the flow of electrons in a circuit. This resistance causes a reduction in current and an increase in voltage in the circuit. In order words, as resistance increases, the current decreases and voltage increases.

This was further stated by Ohm's law that states that as long as the resistance in a wire/conductor remains constant, the voltage across it is directly proportional to the current flowing through a conductor. i.e

V = IR

Where;

R = constant called resistance

I = current flowing through the wire

V = voltage across the wire

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Answer:

The radius of the dish is $R = 18cm$

Explanation:

From the question we are told that

The radius of the orbit is = $R = 35,000km = 35,000 *10^3 m$

The power output of the power is $P = 1 kW = 1000W$

The electric vector amplitude is given as $E = 0.1 mV/m = 0.1 *10^{-3}V/m$

The area of thereciever is $A_R = 5cm^2$

Generally the intensity of the dish is mathematically represented as

$I = \frac{P}{A}$

Where A is the area orbit which is a sphere so this is obtained as

$A = 4 \pi r^2$

$= (4 * 3.142 * (35,000 *10^3)^2)$

$=1.5395*10^{16} m^2$

Then substituting into the equation for intensity

$I_s = \frac{1000}{1.5395*10^{16}}$

$= 6.5*10^ {-14}W/m2$

Now the intensity received by the dish can be mathematically evaluated as

$I_d = \frac{1}{2} * c \epsilon_o E_D ^2$

Where c is thesped of light with a constant value $c = 3.0*10^8 m/s$

$\epsilon_o$ is the permitivity of free space with a value $8.85*10^{-12} N/m$

$E_D$ is the electric filed on the dish

So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

Now making the eletric field intensity the subject of the formula

$E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }$

substituting values

$E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }$

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The incident power on the dish is what is been reflected to the receiver

$P_D = P_R$

Where $P_D$ is the power incident on the dish which is mathematically represented as

$P_D = I_d A_d$

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And $P_R$ is the power incident on the dish which is mathematically represented as

$P_R = I_R A_R$

$= \frac{1}{2} c \epsilon_o E_R^2 A_R$

Now equating the two

$\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R$

Making R the subject we have

$R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }$

Substituting values

$R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }$

$R = 18cm$

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