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lilavasa [31]
2 years ago
10

Transformers: an ideal step-up transformer doubles a primary voltage of 110 v. what is the ratio of the number of turns in its p

rimary coil to the number of turns in the secondary coil?
Physics
1 answer:
Bogdan [553]2 years ago
6 0

The ratio of the number of turns in the primary and the secondary coil of a step-up transformer is 1 : 200.

The formula for calculating the secondary voltage in a transformer is: N = V = Where No and N are the primary and secondary coils' respective turn counts; V and V are the RMS voltages across the primary and secondary coils, and I and Is are the primary and secondary currents.

Calculation: Due to this:

The coil's N2 to Ns turn ratio is 1/20.

We understand that N=1/20 = 4/LP: Is = 20:1.

The transformer uses the following principles: A wire's magnetic field moves as a result of a changing current flowing through it. A wire exposed to a moving magnetic field will create a changing current.One of the coils produces a magnetic field that grows and contracts in response to the changing voltage of the AC when an AC source is connected to it.

In other words, the coil generates a growing magnetic field as the voltage across it rises. The magnetic field generated around the capacitor increases when the voltage reaches its maximum and starts to decline.

learn more about Transformers: brainly.com/question/14184068

#SPJ4

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3 years ago
Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to
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Answer:

KE=1.2036\times 10^{-12}\ J

Explanation:

Given:

  • charge on the alpha particle, q=2e=3.2\times 10^{-19}\ C
  • mass of the alpha particle, m=6.64\times 10^{-27}\ kg
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<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

q.v.B=m.\frac{v^2}{r}

m.v=q.B.r

where:

v = velocity of the alpha particle

v=\frac{q.B.r}{m}

v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}

v=1.2048\times 10^{7}\ m.s^{-1}

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m

m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }

m'=6.6533\times10^{-27}\ kg

now kinetic energy:

KE=m'.c-m.c

KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2

KE=1.2036\times 10^{-12}\ J

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