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beks73 [17]
3 years ago
11

A disk drive plugged into a 120V outlet operates on a voltage of 12V . The transformer that powers the disk drive has 125 turns

on its primary coil.
Part A) Should the number of turns on the secondary coil be greater than or less than 125? greater than or less than
Part B) Explain
Part C) Find the number of turns on the secondary coil.
Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

(A-) Number of turns in secondary coil will be<u> less than 125.</u>

(B-) EXPLAIN- The voltage is to be lowered.<u> The voltage will be reduced from 120 to 9 volts.</u> As a result, a<u> step down transformer </u>will be used, indicating that the secondary coil's number of turns would be less than 125.

(C-)<u> The number of turns on the secondary coil is 12.5</u>

<u></u>

Explanation:

  • STEP DOWN TRANSFORMER - A step-down transformer transforms high voltage (HV) and low current from the transformer's primary side to low voltage (LV) and high current on the secondary side. A phase down transformer is the opposite of this.

In electrical networks and transmission lines, a step-down transformer can be used in a number of ways.

<u>Step down transformer formula-</u>

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

We are given with the -

Primary voltage V_p = 120V

Secondary voltage Vs=12V

Number of primary turns N_p=125

  • <u>To calculate the number of turns on the secondary coil -</u>

       Since , the transformer ratio is given as -

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

So, the number of turns on the secondary coil is - N_s=N_p\times\frac{V_p}{V_s}

            Putting , the given values -:

                125\times\frac{12}{120}

         =   12.5

<u>Therefore , the number of turns on secondary coil is 12.5 .</u>

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Explanation:

We have the following information:

h_{Hg-TOP}=675mmHg=0.675m the barometric reading at the top of the building

h_{Hg-BOT}=695mmHg=0.695m the barometric reading at the bottom of the building

\rho _{air}=1.18 kg/m^{3} density of air

\rho _{Hg}=13600 kg/m^{3} density of mercury

g=9.8/m^{2} gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building P_{TOP} and the perssure at the bottom P_{BOT}:

P_{TOP}=\rho _{Hg} g h_{Hg-TOP} (1)

P_{BOT}=\rho _{Hg} g h_{Hg-BOT} (2)

From (1): P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa (3)

From (2): P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure \Delta P:

\Delta P=P_{BOT} - P_{TOP} (5)

\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa (6)

On the other hand, we have a column of air with a cross-section area A and the same height of the building, lets name it h_{air}.

As pressure is defined as the force F exerted on a specific area A, we can write:

\Delta P=\frac{F}{A} (7)

If we isolate F we have:

F= A \Delta P (8)

Also, the force gravity exerts on this column of air (its weight) is:

F=m_{air} g (9)

Knowing the density of air is: \rho_{air}=\frac{m_{air}}{V_{air}} (10)

where the volume of air can be written as: V_{air}=(A)(h_{air}) (11)

Substituting (1) in (10):

\rho_{air}=\frac{m_{air}}{(A)(h_{air}} (12)

Isolating m_{air}:

m_{air}=(\rho_{air}) (A) (h_{air}) (13)

Substituting (13) in (9):

F=(\rho_{air}) (A) (h_{air}) (g) (14)

Matching (8) and (14)

A \Delta P=(\rho_{air}) (A) (h_{air}) (g) (15)

Isolating h_{air}:

h_{air}=\frac{\Delta P}{g \rho_{air}} (16)

Substituting the known and calculated values:

h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})} (17)

Finally:

h_{air}=230.50 m This is the height of the building

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