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Serjik [45]
3 years ago
13

A neighborhood transformer on the top of a utility pole transforms 12.0 kV 60.0 Hz alternating voltage down to 120.0 V to be use

d inside a house. If the secondary coil of the transformer has 134.0 turns, then how many turns does the primary coil have
Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

<em>The primary coil has 13,400 turns</em>

Explanation:

<u>Voltage Transformers</u>

A transformer is an electrical apparatus that converts an alternating electrical voltage to another. Step-down transformers lower the voltage from higher levels (kilovolts) to consumer levels (120/240 Volts).

The ratio between both voltages can be computed as

\displaystyle r=\frac{V_1}{V_2}

Where V1 is the primary voltage and V2 is the secondary voltage. This ratio depends on the turns ratio of the coils wounded in a common magnetic core.

\displaystyle r=\frac{N_1}{N_2}

Being N1 the number of turns of the coils of the primary side and N2 the number of turns in the secondary coil. Both relations give us

\displaystyle \frac{N_1}{N_2}=\frac{V_1}{V_2}

Solving for N1

\displaystyle N_1=\frac{V_1}{V_2}\cdot N_2

We have:

V_1=12,000\ V\\V_2=120\ V\\N_2=134

Calculate N1

\displaystyle N_1=\frac{12,000}{120}\cdot 134=13,400

The primary coil has 13,400 turns

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A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant
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The work done to stretch the spring will be 112 J.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

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\rm W= \frac{1}{2} kx^2 \\\\ W=\frac{1}{2} \times 350 \times (0.850-0.050)^2 \\\\ W=0.5 \times 350 \times (0.80)^2 \\\\W=112 \ J

To learn more about the spring force refer to the link;

brainly.com/question/4291098

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A wire loop of radius 0.50 m lies so that an external magnetic field of magnitude 0.40 T is perpendicular to the loop. The field
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Make a substitution:

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Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:

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Make another substitution:

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Given values:

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Plug in and solve for ℰ:

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