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Tomtit [17]
3 years ago
10

The current through a 0.1 Henrys (H) inductor is i(t) = 10 t e^-5tA. Find the voltage across the inductor.

Engineering
1 answer:
Lady bird [3.3K]3 years ago
5 0

Answer: d. v(t)=(1-5t)e^{-5t}V

Explanation: <u>Inductance</u> is a property of an inductor: when there is a change in current passing through a conductor, it creates a voltage in the conductor itself and in the other conductors. Inductance unit is Ω.s or henry (H)

So, the relation between Voltage and Current in an inductor is given by

v=L\frac{di}{dt}

in which

L is inductance in H

i is current in A

Current is i(t)=10te^{-5t}, so, derivative will be:

\frac{di}{dt}=10e^{-5t}+10t(-5)e^{-5t}

\frac{di}{dt}=10e^{-5t}-50te^{-5t}

\frac{di}{dt}=10e^{-5t}(1-5t)

Then, voltage is

v=0.1.10.e^{-5t}(1-5t)

v=(1-5t)e^{-5t}

<u>Voltage across the 0.1H inductor is </u>(1-5t)e^{-5t}<u> V</u>

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The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in t
vovikov84 [41]

Answer:

Explanation:

r_2=∞

q=4\pi kT_1(T_2-T_1)\\

q=2\pi kD.ΔT--------(1)

q=hA ΔT=4\pi r_1^2(T_2_s-T_1_s)\\

N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9}  }  ------(3)

By equation (1) and (2)

2\pi kD.ΔT=h.4\pi r_1^2ΔT

2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\-------(4)

From equation (3) and (4)

So for sphere R_a→0

6 0
3 years ago
This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons
telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, d_{prop} = m/sec (s)

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet,  d_{trans} = L/R

(c)

As we know, end-to-end delay or total no delay,

\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}

Here,  $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }

(d)

The expression, time time $t=d_{\text {trans }}$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at time $t=d_{\text {trans }}$ the last bit of the packet has been pushed out or transmitted.

(e)

If \ d_{prop} >d_{trans}

Then, at time $t=d_{\text {trans }}$ the bit has been transmitted from host A, but to condition (1),  the first bit has not reached B.

(f)

If \ d_{prop}

Then, at time $t=d_{\text {trans }}$, the first bit has reached destination on B

Here,s=2.5 \times 10^{8} \mathrm{sec}

\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}

It's given that \ d_{prop} =d_{trans}

Hence,

        \begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}

5 0
3 years ago
More discussion about seriesConnect(Ohm) function In your main(), first, construct the first circuit object, called ckt1, using
STALIN [3.7K]

Answer:

resistor.h

//circuit class template

#ifndef TEST_H

#define TEST_H

#include<iostream>

#include<string>

#include<vector>

#include<cstdlib>

#include<ctime>

#include<cmath>

using namespace std;

//Node for a resistor

struct node {

  string name;

  double resistance;

  double voltage_across;

  double power_across;

};

//Create a class Ohms

class Ohms {

//Attributes of class

private:

  vector<node> resistors;

  double voltage;

  double current;

//Member functions

public:

  //Default constructor

  Ohms();

  //Parameterized constructor

  Ohms(double);

  //Mutator for volatage

  void setVoltage(double);

  //Set a resistance

  bool setOneResistance(string, double);

  //Accessor for voltage

  double getVoltage();

  //Accessor for current

  double getCurrent();

  //Accessor for a resistor

  vector<node> getNode();

  //Sum of resistance

  double sumResist();

  //Calculate current

  bool calcCurrent();

  //Calculate voltage across

  bool calcVoltageAcross();

  //Calculate power across

  bool calcPowerAcross();

  //Calculate total power

  double calcTotalPower();

  //Display total

  void displayTotal();

  //Series connect check

  bool seriesConnect(Ohms);

  //Series connect check

  bool seriesConnect(vector<Ohms>&);

  //Overload operator

  bool operator<(Ohms);

};

#endif // !TEST_H

resistor.cpp

//Implementation of resistor.h

#include "resistor.h"

//Default constructor,set voltage 0

Ohms::Ohms() {

  voltage = 0;

}

//Parameterized constructor, set voltage as passed voltage

Ohms::Ohms(double volt) {

  voltage = volt;

}

//Mutator for volatage,set voltage as passed voltage

void Ohms::setVoltage(double volt) {

  voltage = volt;

}

//Set a resistance

bool Ohms::setOneResistance(string name, double resistance) {

  if (resistance <= 0){

      return false;

  }

  node n;

  n.name = name;

  n.resistance = resistance;

  resistors.push_back(n);

  return true;

}

//Accessor for voltage

double Ohms::getVoltage() {

  return voltage;

}

//Accessor for current

double Ohms::getCurrent() {

  return current;

}

//Accessor for a resistor

vector<node> Ohms::getNode() {

  return resistors;

}

//Sum of resistance

double Ohms::sumResist() {

  double total = 0;

  for (int i = 0; i < resistors.size(); i++) {

      total += resistors[i].resistance;

  }

  return total;

}

//Calculate current

bool Ohms::calcCurrent() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  current = voltage / sumResist();

  return true;

}

//Calculate voltage across

bool Ohms::calcVoltageAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double voltAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      voltAcross += resistors[i].voltage_across;

  }

  return true;

}

//Calculate power across

bool Ohms::calcPowerAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double powerAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      powerAcross += resistors[i].power_across;

  }

  return true;

}

//Calculate total power

double Ohms::calcTotalPower() {

  calcCurrent();

  return voltage * current;

}

//Display total

void Ohms::displayTotal() {

  for (int i = 0; i < resistors.size(); i++) {

      cout << "ResistorName: " << resistors[i].name << ", Resistance: " << resistors[i].resistance

          << ", Voltage_Across: " << resistors[i].voltage_across << ", Power_Across: " << resistors[i].power_across << endl;

  }

}

//Series connect check

bool Ohms::seriesConnect(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  vector<node> temp = ohms.getNode();

  for (int i = 0; i < temp.size(); i++) {

      this->resistors.push_back(temp[i]);

  }

  return true;

}

//Series connect check

bool Ohms::seriesConnect(vector<Ohms>&ohms) {

  if (ohms.size() == 0) {

      return false;

  }

  for (int i = 0; i < ohms.size(); i++) {

      this->seriesConnect(ohms[i]);

  }

  return true;

}

//Overload operator

bool Ohms::operator<(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  if (this->sumResist() < ohms.sumResist()) {

      return true;

  }

  return false;

}

main.cpp

#include "resistor.h"

int main()

{

   //Set circuit voltage

  Ohms ckt1(100);

  //Loop to set resistors in circuit

  int i = 0;

  string name;

  double resistance;

  while (i < 3) {

      cout << "Enter resistor name: ";

      cin >> name;

      cout << "Enter resistance of circuit: ";

      cin >> resistance;

      //Set one resistance

      ckt1.setOneResistance(name, resistance);

      cin.ignore();

      i++;

  }

  //calculate totalpower and power consumption

  cout << "Total power consumption = " << ckt1.calcTotalPower() << endl;

  return 0;

}

Output

Enter resistor name: R1

Enter resistance of circuit: 2.5

Enter resistor name: R2

Enter resistance of circuit: 1.6

Enter resistor name: R3

Enter resistance of circuit: 1.2

Total power consumption = 1886.79

Explanation:

Note

Please add all member function details.Its difficult to figure out what each function meant to be.

8 0
3 years ago
Consider a turbojet powered airplane flying at a standard altitude of 30,000ft at a velocity of 500 mph. The turbojet engine its
grandymaker [24]

Answer:

T  = 20.42 N

Explanation:

given data

standard altitude = 30,000 ft

velocity Ca = 500 mph = 0.4 m/s

inlet areas Aa = 7 ft² = 0.65 m²

exit areas Aj = 4.5 ft²  =  0.42 m²

velocity at exit Cj = 1600 ft/s = 487.68 m/s

pressure exit \rhoj = 640 lb/ft²   = 0.3 bar

solution

we get here thrust of the turbojet that is  express as

thrust of the turbojet T = Mg × Cj - Ma × Ca + ( \rhoj Aj - \rhoa Ag )   .............1

here Ma = Mg

Ma = \rhoa × Ca Aa = 0.042 kg/s

put value in equation 1 we get

T = 0.042 × (487.68 -0.14) + ( 0.3 ×  - 0.3 × 0.65 )  

T  = 20.42 N

5 0
3 years ago
Inhalation is the most common way for a(n)
Mademuasel [1]

inhalation is the most common way for chemical to enter the body

6 0
3 years ago
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