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2 years ago

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The distribution circuit of a residential power line is operated at 2000 V rms. This voltage must be reduced to 240 V rms for us

e within the residences. If the secondary side of the transformer has 475 turns, how many turns are in the primary?

1 answer:

forsale [732]2 years ago

7 0

Answer:

$N_p\approx3958$

Explanation:

In an ideal transformer the relationship between the voltages is proportional to the ratio between the number of turns of the windings. Thus:

$\frac{V_p}{V_s} =\frac{N_p}{N_s}$

Where:

$V_p=Voltage\hspace{3}in\hspace{3}primary\hspace{3}coil$

$V_s=Voltage\hspace{3}in\hspace{3}secondary\hspace{3}coil$

$N_p=Turns\hspace{3}on\hspace{3}primary\hspace{3}coil$

$N_s=Turns\hspace{3}on\hspace{3}secondary\hspace{3}coil$

So, solving for $N_p$

$N_p=N_s*\frac{V_p}{V_s} =475*\frac{2000}{240} =3958.333333\approx3958$

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

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A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th

fiasKO [112]

Let's use the mirror equation to solve the problem:
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where f is the focal length of the mirror, $d_o$ the distance of the object from the mirror, and $d_i$ the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for $d_i$ by using the numbers given in the text of the problem:
$\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}$
$\frac{1}{d_i}= -\frac{7}{60 cm}$
$d_i = -8.6 cm$
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Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

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We can write it as follows :

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