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Inessa05 [86]
3 years ago
8

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

Physics
2 answers:
Nesterboy [21]3 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]3 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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Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.
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Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  D = 0.59 \  m    

Explanation:

From the question we are told that

      The best resolution is  \theta  =  0.3 \  arcsecond

       The  wavelength is  \lambda  =  700 \  nm =  700 *10^{-9 } \  m

       

Generally the

         1 arcminute  = >  60 arcseconds

=>      x arcminute =>   0.3 arcsecond

So

       x =  \frac{0.3}{60 }

=>    x = 0.005 \  arcminutes

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       z =  \frac{0.005}{60 }

=>      z =  8.333 *10^{-5}  \ degree

Converting to radian  

           \theta  = z =  8.333 *10^{-5}  * 0.01745 = 1.454 *10^{-6} \  radian

Generally the resolution is mathematically represented as

            \theta  =  \frac{1.22 *  \lambda  }{ D}

=>    D =  \frac{1.22 * \lambda }{\theta }

=>     D =  \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }    

=>     D = 0.59 \  m    

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Answer:

Part(a): The equation of motion is \bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}.

Part(b): The equation of motion is  \bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}.

Explanation:

If 'm' be the mass of the object, 'k' be the force constant and '\beta' be the damping constant, then the equation of motion of the particle can be written as

\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)

Given m = 1 Kg, k = 14 N~m^{-1}, \beta = 9. Substituting these values in equation (I),

\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0

Taking a trial solution x(t) = e^{mt}, the auxiliary equation can be written as

m^{2} + 9m + 14 = 0............................................................(II)

and its solutions are m_{1} = -2~and~m_{2} = -7, resulting the general solution

x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)

The velocity at any instant of time of the mass is

v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)

Part(a):

Given x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}. Substituting these values in equation (III) and (IV),

&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)

Solving equations (V) and (VI), we have

C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}

So the equation of motion is

x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}

Part(b):

Given x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}. Substituting these values in equation (III) and (IV),

&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)

Solving equations (V) and (VI), we have

C_{1} = -1~and~C_{2} = \dfrac{11}{5}

So the equation of motion is

x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}

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