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Inessa05 [86]
4 years ago
8

A step-down transformer has more loops on which coil?

Physics
2 answers:
Nesterboy [21]4 years ago
6 0
A step-down transformer has more loops in :  A. Primary coil

Primary coil refers to the coil to which alternating voltage is supplied. It's usually connected to the AC supply

hope this helps
MrRissso [65]4 years ago
3 0

Answer:

A). Primary coil

Explanation:

As we know that in an ideal transformer the flux linked with primary coil and secondary coil for each turn must be same

so we can say that

\phi _p = N_p \phi_0

\phi_s = N_s\phi_0

so as per faraday's law we can say that two EMF is given as

E_s = N_s\frac{d\phi_0}{dt}

E_p = N_pfrac{d\phi_0}{dt}

now we have

\frac{E_s}{E_p} = \frac{N_s}{N_p}

so we can say that

for step down transformer the output voltage is less than the input transformer

E_s < E_p

so we will have

N_s < N_p

so turns will be more in

A) Primary coil

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Answer:

Explanation:

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v_x=u_x+a_x\times t

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Earth's gravitational force just got three times stronger! What happens to your weight?
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A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz,
masha68 [24]

Answer:

a) The car was moving at a speed of 29.167\ m.s^{-1}

b) The negative sign of v_o denotes that the observer is coming towards the police car which is the source of the sound.

c) f_o=1283.33\ Hz

Explanation:

Given:

  • original frequency of the source, f=1200\ Hz
  • speed of the source, v_s=0\ m.s^{-1}
  • velocity of the obstacle car be, v_o
  • speed of sound, s=350\ m.s^{-1}
  • observed frequency, f_o=1100\ Hz

<u>Using the equation from the Doppler's effect:</u>

\frac{f_o}{f} =\frac{(s+v_o)}{(s-v_s)}

\frac{1100}{1200} =\frac{(350+v_o)}{350-0}

v_o=-29.167\ m.s^{-1}

a)

The car was moving at a speed of 29.167\ m.s^{-1}

b)

The negative sign of v_o denotes that the observer is coming towards the police car which is the source of the sound.

c)

Now when, v_s=50\ m.s^{-1}

Then, f_o=?

Using the Doppler's eq.:

\frac{f_o}{1200} =\frac{(350+(-29.167))}{(350-50)}

f_o=1283.33\ Hz

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