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Oxana [17]
3 years ago
9

Suppose a large power plant generates electricity at 12.0 kV. Its old transformer once converted this voltage to 315 kV. The sec

ondary coil of this transformer is being replaced so that its output can be 730 kV for more efficient cross-country transmission on upgraded transmission lines.
Randomized Variables

Vi = 315 kV

V2 = 730 kV


(a) What is the ratio of turns in the new secondary to the number of turns in the old secondary?

(b) What is the ratio of new current output to the old current output for the same power input to the transformer?
Physics
1 answer:
SpyIntel [72]3 years ago
7 0

Answer:

  • 2.32
  • 0.43

Explanation:

12.0 kv primary voltage

315 kv secondary voltage ( converted voltage ) V1 or Vo

v2 (Vn)= 730 kv new secondary voltage

a) Ratio of turns in 730 kv to turns in 315 kv

\frac{Vn}{Vo} = \frac{Nn}{No} = \frac{730}{315}  therefore the ratio of turns = 2.317 ≈ 2.32

B) ratio of the new current output to the old current output for the same power input to the transformer

since the power input is the same

\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }     equation 1

Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current

therefore equation 1 becomes

\frac{In}{Io}  = \frac{Vo}{Vn} =  315 / 730 = 0.43

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The function x = (5.2 m) cos[(5πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 5.3 s, what are the (a) di
gulaghasi [49]

Answer:

(a) Displacement = - 3.0576 m

(b) Velocity  =-66.48 m/s

(c)Acceleration   = -753.39 m²/s

(d)The phase motion is 26.7 \pi.

(e)Frequency =2.5 Hz.

(f)Time period =0.4 s

Explanation:

Given function is

x= (5.2 m)cos[ (5\pi \  rad/s)t+ \frac\pi5]

(a)

The displacement includes the parameter t, so,at time t=5.3 s

x|_{t=5.3}= (5.2 m)cos[ (5\pi \  rad/s)5.3+ \frac\pi5]

           = (5.2 m)cos[ 26.5\pi+ \frac\pi5]

           =(5.2)(-0.588)m

           = - 3.0576 m

(b)

x= (5.2 m)cos[ (5\pi \  rad/s)t+ \frac\pi5]

To find the velocity of simple harmonic motion, we need to find out the first order derivative of the function.

v=\frac{dx}{dt}

 =\frac{d}{dt} (5.2 m)cos[ (5\pi \  rad/s)t+ \frac\pi5]

  =  (5.2 m)(-5\pi)sin[ (5\pi \  rad/s)t+ \frac\pi5]

  =  -26\pi sin[ (5\pi \  rad/s)t+ \frac\pi5]

Now we can plug our value t=5.3 into the above equation

v=  -26\pi sin[ (5\pi \  rad/s)5.3\ s+ \frac\pi5]

 =-66.48 m/s

(c)

To find the acceleration of simple harmonic motion, we need to find out the second order derivative of the function.

v=  -26\pi sin[ (5\pi \  rad/s)t+ \frac\pi5]

a=\frac{d^2x}{dt^2}

 =\frac{dv}{dt}

 =\frac{d}{dt}(  -26\pi sin[ (5\pi \  rad/s)t+ \frac\pi5])

 =  -26\pi (5\pi)cos[ (5\pi \  rad/s)t+ \frac\pi5]

 =  -130\pi^2cos[ (5\pi \  rad/s)t+ \frac\pi5]

Now we can plug our value t=5.3 into the above equation

a=  -130\pi^2cos[ (5\pi \  rad/s)5.3 \ s+ \frac\pi5]

  = -753.39 m²/s

(d)

The general equation of SHM is

x=x_mcos(\omega t+\phi)

x_m is amplitude of the displacement, (\omega t+\phi) is phase of motion, \phi is phase constant.

So,

(\omega t+\phi)=5\pi t+\frac\pi5

Now plugging t=5.3s

(\omega t+\phi)=5\pi \times 5.3+\frac\pi5

             =26.7 \pi

The phase motion is 26.7 \pi.

The angular frequency \omega = 5\pi

(e)

The relation between angular frequency and frequency is

\omega =2\pi f

\therefore f=\frac{\omega}{2\pi}

     =\frac{5\pi}{2\pi}

    =\frac52

   = 2.5 Hz

Frequency =2.5 Hz.

(f)

The relation between frequency and time period is

T=\frac1 f

   =\frac1{2.5}

  =0.4 s

Time period =0.4 s

6 0
3 years ago
A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the
Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

V = 0

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr

V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr

V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}

V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})

V_{r}=-0.00012190\ V

V_{r}=-1.219\times10^{-4}\ V

The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). We need to calculate the potential at a distance r = R

Using formula of  potential difference

V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}

V_{R}=-0.0003759\ V

V_{R}=-3.759\times10^{-4}\ V

The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Hence, This is the required solution.

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