Question 9 of 25

A full-adder is a combinational circuit that forms the arithmetic sum of three input bits.

Vinvika [58]

(b) correct it is false

5 0

2 years ago

P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si

klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V 0.184 W

2.499 mV 125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

$V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\$ *5

= 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:

V_o = A_voc*V_i*(R_o/R_l+R_o)

= 4.545*(100/100+50)

= 3.03 V

Now we can determine delivered power:

P_L = V_o^2/R_L

= 0.184 W

Apply voltage-divider principle to the circuit:

V_o = (R_o/R_o+R_s)*V_s

= 50/50+100*10^3*5

= 2.499 mV

Now we can determine delivered power:

P_l = V_o^2/R_l

= 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.

4 0

2 years ago

Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes

Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

At the inlet of turbine P=10 MPa ,T=500 C

AT the exit of turbine P=10 KPa ,x=0.9

Required power=5 MW

From steam table

At 10 MPa and 500 C:

h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)

At 10 KPa:

$h_g$ =2675.1 KJ/Kg, $h_f$ =417.51 KJ/Kg

$s_g$ = 7.3 KJ/Kg-K , $s_f$ =1.3 KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= $h_f+x(h_g- h_f)$

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34 KJ/Kg

Lets take m is the mass flow rate of steam

So $5\times 10^3=m\times (3374-2449.34)$

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0

3 years ago

Calculate the theoretical density of FCC iron (eg. austenitic stainless steel). The lattice parameter for FCC iron is 0.357 nm a

Ann [662]

Answer: 12.4 feet

Explanation:

If there is a smooth transition and there is no change in slopes, energy considerations can be used

The cube has a kinetic energy of

ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2

At the highest point when there is a gain in potential energy

pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

If there is no loss in energies,

pe = ke

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

h = 2000 /322 = 6.211 (ft)

= h / sin(30) = 12.4 ft

8 0

2 years ago

As Energy is added to ice, what happens to the temperature over time and why?

Reil [10]

As ice melts into water, kinetic energy is being added to the particles. This causes them to be 'excited' and they break the bonds that hold them together as a solid, resulting in a change of state: solid -> liquid.

5 0

2 years ago

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