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arlik [135]
2 years ago
5

A circuit contains three resistors connected in parallel. The value of R1 is 2 K , the value or R2 is 6 K , and the value of R3

is 10 K . If the supplied voltage is 100 VDC, what is the total power in the circuit
Computers and Technology
1 answer:
mezya [45]2 years ago
6 0
In a parallel connection, the voltage is same in every branch.
Now, three <span>three resistors connected in parallel.
R1 = 2k ohm
</span>R2 = 6k ohm
R3 = 10k ohm
in parallel, net resisitance  = \frac{ R_{1}R_{2}R_{3} }{R_{1}R_{2} + R_{2}R_{3} + R_{3}R_{1}}

Now, putting the values, we get, R net = 1.30 k ohm.
Ans, voltage = 100 VDC
Thus, power = \frac{ V^{2} }{R} = \frac{ 100^{2} }{1.3 k} \\ \\ where \\ k = 10^{3} 
                      = 7.69 Watt
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George and Miguel are considering opening up a shoe store but first need to do market research. Which one of these is NOT part o
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The correct answer is D, Draw conclusions and make decisions for their business based on the research results.
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2 years ago
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Write a program for determining if a year is a leap year. In the Gregorian calendar system you can check if it is a leaper if it
SOVA2 [1]

Answer:

def leap_year_check(year):

return if int(year) % 4 == 0 and (int(year) % 100 != 0 or int(year) % 400 == 0)

Explanation:

The function is named leap_year_check and takes in an argument which is the year which we wish to determine if it's a new year or not.

int ensures the argument is read as an integer and not a float.

The % obtains the value of the remainder after a division exercise. A remainder of 0 means number is divisible by the quotient and a remainder other wise means it is not divisible by the quotient.

If the conditions is met, that is, (the first condition is true and either the second or Third condition is true)

the function leap_year_check returns a boolean ; true and false if otherwise.

8 0
2 years ago
____ twisted pair is the least quality twisted pair wire that should be used in a data/voice application.
xxMikexx [17]

Answer:

Category 5 (Cat 5) twisted pair is the least quality twisted pair wire that should be used.

Explanation:

The reason is as follows:

  1. Category 1 twisted pair is not suitable for transmitting data but comes in handy for telephone communications. Hence, it should not be used in a data/voice application.
  2. Category 2 twisted pair is suitable for data transmission, however, it can only transmit at 4Mbps which is very slow. Hence, it should not be used in a data/voice application.
  3. Category 3 twisted pair can transmit data with speed reaching 10Mbps which is slow for voice/data communication.
  4. Category 4 twisted pair can transmit data with speed reaching 16Mbps and is used in token ring networks. It is slow for communicating in a voice/data application.
  5. Category 5 twisted pair is the least qualified to be used in a voice/data application because it can transmit data at speeds that reach 100Mbps.

Hence, Cat 5 or Category 5 twisted pair is the least quality twisted pair that should be used in a data/voice application.

6 0
2 years ago
Modify the TimeSpan class from Chapter 8 to include a compareTo method that compares time spans by their length. A time span tha
My name is Ann [436]

Answer:

Check the explanation

Explanation:

Here is the modified code for

TimeSpan.java and TimeSpanClient.java.

// TimeSpan.java

//implemented Comparable interface, which has the compareTo method

//and can be used to sort a list or array of TimeSpan objects without

//using a Comparator

public class TimeSpan implements Comparable<TimeSpan> {

   private int totalMinutes;

   // Constructs a time span with the given interval.

   // pre: hours >= 0 && minutes >= 0

   public TimeSpan(int hours, int minutes) {

        totalMinutes = 0;

        add(hours, minutes);

   }

   // Adds the given interval to this time span.

   // pre: hours >= 0 && minutes >= 0

   public void add(int hours, int minutes) {

        totalMinutes += 60 * hours + minutes;

   }

   // Returns a String for this time span such as "6h15m".

   public String toString() {

        return (totalMinutes / 60) + "h" + (totalMinutes % 60) + "m";

   }

   // method to compare this time span with other

   // returns a negative value if this time span is shorter than other

   // returns 0 if both have same duration

   // returns a positive value if this time span is longer than other

   public int compareTo(TimeSpan other) {

        if (this.totalMinutes < other.totalMinutes) {

            return -1; // this < other

        } else if (this.totalMinutes > other.totalMinutes) {

            return 1; // this > other

        } else {

            return 0; // this = other

        }

   }

}

// TimeSpanClient.java

public class TimeSpanClient {

   public static void main(String[] args) {

        int h1 = 13, m1 = 30;

        TimeSpan t1 = new TimeSpan(h1, m1);

        System.out.println("New object t1: " + t1);

        h1 = 3;

        m1 = 40;

        System.out.println("Adding " + h1 + " hours, " + m1 + " minutes to t1");

        t1.add(h1, m1);

        System.out.println("New t1 state: " + t1);

        // creating another TimeSpan object, testing compareTo method using the

        // two objects

        TimeSpan t2 = new TimeSpan(10, 20);

        System.out.println("New object t2: " + t2);

        System.out.println("t1.compareTo(t2): " + t1.compareTo(t2));

        System.out.println("t2.compareTo(t1): " + t2.compareTo(t1));

        System.out.println("t1.compareTo(t1): " + t1.compareTo(t1));

   }

}

/*OUTPUT*/

New object t1: 13h30m

Adding 3 hours, 40 minutes to t1

New t1 state: 17h10m

New object t2: 10h20m

t1.compareTo(t2): 1

t2.compareTo(t1): -1

t1.compareTo(t1): 0

4 0
2 years ago
Write the definition of a function named quadratic that receives three double parameters a, b, c. If the value of a is 0 then th
VMariaS [17]

Answer:

#include <iostream>

#include <cmath>

using namespace std;

//initialize function quadratic

void quadratic(double, double, double);

int main() {

   //declare double variables a, b and c

   double a,b,c;

   //take input from user

   cin>>a>>b>>c;

   //call function quadratic

   quadratic(a,b,c);

return 0;

}

void quadratic(double a, double b, double  c){

   double root,n;

   //check if variable a is equal to zero

   if(a==0){

       cout<<"no solution for a=0"<<endl;

       return;

   }

   //check if b squared - 4ac is less than zero

   else

   if(((b*b)-(4*a*c))<0){

       cout<<"no real solutions"<<endl;

       return;

   }

   //print the largest root if the above conditions are not satisfied

   else{

       n=((b*b)-(4*a*c));

       root=(-b + sqrt(n)) / (2*a);

       cout<<"Largest root is:"<<root<<endl;

   }

   return ;

}

Explanation:

Read three double variables a, b and c from the user and pass it to the function quadratic. Check if the value of variable a is equal to zero. If true, print "no solution for a=0". If this condition is false, check if b squared - 4ac is less than zero. If true, print "no real solutions". If this condition is also false, calculate the largest solution using the following formula:

largest root = (-b + square root of (b squared - 4ac)) / (2*a)

Input 1:

2 5 3

Output 2:

Largest root is:-1

Input 2:

5 6 1

Output 2:

Largest root is:-0.2

5 0
2 years ago
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