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GarryVolchara [31]
3 years ago
10

Find the size of the resistor which would limit the current to a value of 2 mA when connected to an 8 V supply

Engineering
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

R = 4000 ohms

Explanation:

Given that,

Current, I = 2 mA

The voltage of the supply, V = 8 V

We need to find the resistance of the resistor. We know that, Ohm's law is as follows :

V = IR

Where

R is resistance

So,

R=\dfrac{V}{I}\\\\R=\dfrac{8}{2\times 10^{-3}}\\\\\\R=4000\ \Omega

So, the resistance of the resistor is 4000 ohms.

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Answer:

#include<iostream>

#include <iomanip>

using namespace std;

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double getTotal(double [], int);

double getAverage(double [], int);

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double getTotal(int rainFall,double NUM_MONTHS[])

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{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

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                 largest = values[month];

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double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

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int main()

{

double rainFall[NUM_MONTHS];

 for (int month = 0; month < NUM_MONTHS; month++)

  {

     cout << "Enter the rainfall (in inches) for month #";

     cout << (month + 1) << ": ";

     cin >> rainFall[month];

 

     while (rainFall[month] < 0)

     {

      cout << "Rainfall must be 0 or more.\n"

             << "Please re-enter: ";

      cin >> rainFall[month];

     }

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  cout << getTotal(rainFall, NUM_MONTHS)

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}

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