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Bond [772]
3 years ago
5

Which of the following statements are true concerning AC circuit that contains both resistance and inductance? A. The current an

d voltage will be in phase., B. The current will lead the voltage ., C. The current will lag the voltage. D. The voltage will lag the current.
Engineering
1 answer:
maw [93]3 years ago
3 0

The current will lag the voltage in AC circuit that contains both resistance and inductance.

Answer: C

Explanation

There is no inductance only circuits in reality.

The circuits containing inductance has also a lower amount of resistance.

The current flows in both resistance and inductance.

There is a drop in the total voltage in resistance and inductance giving rise to the voltage applied in the coil when connected in a series.

An example being inductance coil an AC circuit connected to both resistance and inductance in series.

From the vector diagram, this conclusion can be drawn.

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A pipeline is a system of connected pipelines that can be either underground or out in the environment. These pipelines are used to transport or distribute water, gas, and oil.

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2 years ago
Determine the number of flipflops required to build a binary counter that count from 0 to 2043
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10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

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3 years ago
How long should the shafts remain in the furnace to achieve a desired centerline temperature of 800K? 2) Determine the temperatu
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Answer:

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6 0
4 years ago
What is the objective of phasing out an INDUCTION MOTOR before putting the machine into commission?
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The main objective of phasing out an INDUCTION MOTOR is to identify the ends of the stator coils.

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An induction motor is a device also known as an asynchronous motor due to its irregular velocity.

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8 0
2 years ago
A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th
Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ    

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

 cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    

b)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

x_A and x_B are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, x_A = 248 /(248 + 42.8) = 0.83

x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

5 0
3 years ago
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