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natima [27]
3 years ago
8

You doubled the voltage frequency in an RL series AC circuit, the inductive resistance would?

Engineering
2 answers:
Vesna [10]3 years ago
7 0

<u>Answer: </u>

When the voltage frequency is doubled, the resistance will remain same in RL series AC circuit.

<u>Explanation: </u>

A circuit that has an inductance L and Resistance R is said to be a RL or LR circuit. The inductance in the circuit is directly proportional to the Frequency F and Inductive Reactance XL of the circuit.  

\mathrm{XL}=2 \times \pi \times \mathrm{F} \times \mathrm{L}

So, whenever there is a change in the frequency it directly changes the inductance and inductive reactance of the circuit but Resistance will remain unaffected.

Resistance in a RL circuit is independent of the frequency. Therefore, in the given case, the resistance will remain the same only the reactance will get doubled.

Ivenika [448]3 years ago
6 0

Answer:

Remain the same.

Explanation:

If you doubled the voltage frequency in an RL series AC circuit, the inductive resistance would <u>remain the same</u>.

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grin007 [14]

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

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V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

5 0
2 years ago
People with skills and training in areas such as marketing or accounting are an important part of the manufacturing industry.
adelina 88 [10]

Answer:

true

Explanation:

8 0
2 years ago
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin
Marina CMI [18]

Solution:

Given that :

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Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$  ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

       $=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

       $=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

       $=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

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7 0
3 years ago
If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

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Explanation:

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As we know that damping ratio given as follows

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Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
g A heat exchanger is designed to is to heat 2,500 kg/h of water from 15 to 80 °C by engine oil. The configuration of the heat e
sergey [27]

Answer:

See explaination

Explanation:

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8 0
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