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aliina [53]
3 years ago
11

Signal generator‘s internal impedance is purely resistive and has an open-circuit voltage of 3.5 V. When the generator is loaded

with 22 ohms of resistance, its terminal voltage drops to 2.8 V. What is the generator’s output impedance (pure resistance)?
Engineering
1 answer:
Bezzdna [24]3 years ago
4 0

Answer:

r = 5.5 ohms

Explanation:

Given:-

- The open circuit voltage, Vo = 3.5 V

- The terminal voltage, Vt = 2.8 V

- The load, R = 22 ohms

- The internal resistance = r

Find:-

What is the generator’s output impedance (pure resistance)?

Solution:-

- We see that the source Voltage (Vo) is not entirely used for the attached load. Some of the source voltage is dropped within due to the source internal resistance or impedance.

- The terminal voltage (Vt) is the amount of Voltage drop across the load. The current drawn by the load I can be determined by Ohm's Law:

                            Vt = I*R

                            I = Vt / R

                            I = 2.8 / 22

                            I = 0.12727 Amps

- Since, the attached load (R) and the pure impedance (r) of the source are in series. The current  ( I ) is constant across both. The potential drop across the pure resistance (r) can be determined from Ohm's law:

                          Vo - Vt = I*r

                          r = ( Vo - Vt ) / I

                          r = ( 3.5 - 2.8 ) / 0.12727

                          r = 5.5 ohms

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Answer:

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Explanation:

Given the data in the question;

To determine whether the aircraft component will fracture, given a fracture toughness of 30 Mpa√m, stress level of 355 and maximum internal crack length of 1.39 mm.

On a similar component, it has been said that fracture results at a stress of 237 MPa when the maximum (or critical) internal crack length is 2.78 mm.

so we first of all solve for the parameter Y in the condition where fracture occurred.

K_{Ic = 30 Mpa√m

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2α = 2.78 mm = 2.78 × 10⁻³ m  

so

Y = K_{Ic / σ√πα

we substitute

Y = (30 Mpa√m) / (237 MPa)√(π(2.78 × 10⁻³ m / 2 ) )

Y =  (30 Mpa) / (237)( 0.06608187 )

Y = 30 / 15.6614

Y = 1.9155

Next we solve for Yσ√πα for the second case;

σ = 355 Mpa, 2α = 1.39 mm = 1.39 × 10⁻³ m

so

Yσ√πα = 1.9155 × 355 Mpa × √( π × (1.39 × 10⁻³ m / 2) )

= 1.9155 × 355 × 0.0467269

= 31.8 Mpa√m

so

( 31.8 Mpa√m ) > K_{Ic ( 30 Mpa√m )

Therefore, fracture will occur since ( 31.8 Mpa√m ) is greater than the K_{Ic of the material ( 30 Mpa√m )

4 0
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Answer:

#include<iostream>

int main() {

  float num_1, num_2, num_3, num_4, average;

  //Taking input for four numbers

  std::cout << "Enter first number(integer or floating-point)";

  std::cin >> num_1;

  std::cout << "Enter second number(integer or floating-point)";

  std::cin >> num_2;

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Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

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a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

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- Evaluate each at time t = 2 s.

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                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

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                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

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