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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:

8 to 10 times

Explanation:

For dry road

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So the de acceleration a = μ g

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a = 0.4 x 10 = 4 m/s ²

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We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

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3 years ago
Discuss in detail the following methods used to redistribute income and wealth in cash grants?​
aev [14]

Answer:

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7 0
3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

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PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

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P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

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