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3 years ago

How are step up transformers used in the transmission of electrical energy

Physics

1 answer:

tensa zangetsu [6.8K]3 years ago

8 0

Answer:

Transformers are used to increase or decrease the voltage of AC currents

Explanation:

A transformer is a device consisting of two coils (called primary and secondary coil) wrapped at the two sides of a soft iron core. When an AC current is present in the primary coil, it induces a magnetic field inside the core, and the presence of this changing magnetic field induces a voltage (and a current) into the secondary coil.

The voltages in the primary and the secondary coil are related by the transformer equation:

$\frac{V_p}{V_s}=\frac{N_p}{N_s}$

where

Vp, Vs are the voltages in the primary and secondary coil

Np, Ns are the number of turns in the primary and secondary coil

There are two types of transformers:

- Step-up transformers: these have $N_s > N_p$ , so that $V_s > V_p$ , which means that they increase the voltage. They are used to increase the voltage of the AC current produced by the power plants, before being sent into the transmission lines.

- Step-down transformers: these have $N_s < N_p$ , so that $V_s < V_p$ , which means that they decrease the voltage. They are used at the end of the transmission lines, before the houses, in order to decrease the voltage and allow the household appliances to work properly (in fact, household appliances need lower voltages to work)

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Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu

morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

$U=mgh$

$U=2.35\ kg\times 9.8\ m/s^2\times 1.3\ m$

U = 29.93 Joules

So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

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3 years ago

5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj

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Answer:

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Explanation:

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3 years ago

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

Natali5045456 [20]

The total displacement of the toy car at the given positions is 0.

The given parameters;

First displacement of the car, = 5 cm left
Second displacement of the car, = 8 cm right
Third displacement of the car, = 3 cm to the left

The total displacement of the car is calculated as follows;

Let the left direction be "negative direction"
Let the right direction be "positive direction"

$\Delta x = - \ 5\ cm \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0$

Thus, the total displacement of the toy car at the given positions is 0.

Learn more about displacement here: brainly.com/question/18158577

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2 years ago

Read 2 more answers

Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th

Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

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4 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

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3 years ago