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Novosadov [1.4K]
3 years ago
9

The primary coil of a transformer contains100 turns; the secondary has 200 turns. The primary coil isconnected to a size-AA batt

ery that supplies a constant voltage of1.5 volts. What voltage would be measured across the secondarycoil? zero
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

There  will be no voltage across  secondary coil because DC voltage source is used in primary coil so there is no electromotive force induced in secondary coil.

Explanation:

In this question we have given

AA battery is used to supplies a constant voltage of 1.5 volts to primary coil of  transformer. In this case, voltage source is DC source which is providing constant voltage and for a transformer to work it is necessary to use an AC source.

Therefore, no EMF will induce in the secondary coil

and we know that in a transformer,

\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1} }=\frac{N_{2} }{N_{1} }...........(1)

hence from above equation it is clear that,

\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1}}............(2)

Here ,

E_{2}=0

Put E_{2}=0 in equation (2)

We got

V_{2}=0

There  will be no voltage across  secondary coil because there is no electromotive force induced in secondary coil.

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Answer:

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Is V directly proportional to I (From V=IR)??<br>Or is V inversely proportional to I (From P=IV) ???
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First of all, I is proportional V according to the Ohm's Law. R is merely a constant you need to obtain an equation. However, it is true that R changes with temperature and pressure, therefore Ohm's Law is only applicable in an invariable environment. Also this constant R is different for different materials.

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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

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b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

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