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Смешанное соединение проводников. Параллельное соединение

Параллельное соединение

Последовательное соединение

Закон Ома

00:00:06

We have solved a lot of problems on calculating

00:00:10

electrical circuits on connecting

00:00:12

conductors. Today is the last lesson on

00:00:14

this topic. Tomorrow we are writing a

00:00:16

test. The topic is

00:00:21

complicated problems

00:00:32

on connecting conductors. Of

00:00:44

course, these won’t be on the test tomorrow,

00:00:47

but since you are still a physical class,

00:00:51

you need to know such things at home.

00:00:55

task summary today you

00:01:00

will get acquainted with some ideas

00:01:02

that I have not talked about before in Kirik

00:01:09

8 8th grade

00:01:12

task number 2 b high page 42

00:01:18

2b high page 42

00:01:23

further 5 high page 44

00:01:35

number six high page 52

00:01:43

number six high

00:01:47

page 55

00:01:50

and number four sufficient level

00:02:00

page 58

00:02:05

prepare for the test

00:02:11

the topic of the test or

00:02:14

electric current Ohm's law you see

00:02:17

here on different pages the problem is

00:02:20

located that is, you will repeat

00:02:22

all the material that you

00:02:24

will need to successfully write

00:02:26

the test and now let's start with problem number eight

00:02:30

thirty-nine of

00:02:32

our pony

00:02:39

let's see what this is a task,

00:02:49

this is a task for the curious, I have no

00:02:52

doubt that you are one of those

00:02:55

lyceum students, so let’s deal with a

00:02:57

homogeneous wire with a resistance of 360 m,

00:03:01

connected with a ring at which points of this

00:03:05

ring you need to connect to it in order to

00:03:08

get a resistance of 80 tons, a

00:03:11

homogeneous wire with a resistance of 360 m,

00:03:14

let’s designate this resistance as r

00:03:17

zero 360 is connected by a ring at what

00:03:27

points of this ring you need to connect to it

00:03:29

to get a

00:03:30

resistance of 80 m, let's let this

00:03:33

resistance be designated p small

00:03:36

80th at what points of the ring you need to connect to it let

00:03:40

's think about

00:03:43

what we will look for so you have a piece of

00:03:47

wire, for example, like this If we had a

00:03:49

Rio Chord, you would roll a ring out of it,

00:03:52

if this ring is broken, then the

00:03:55

resistance of the wire will be 360. We rolled a

00:03:58

ring out of it, here it is at which points of

00:04:04

the ring you need to connect from 1. here's the

00:04:09

second one.

00:04:10

for example, and we connected a device to it,

00:04:14

well, let’s say it’s meter by meter, it’s a device for

00:04:17

measuring resistance at which points of the

00:04:19

ring you need to connect to get

00:04:22

the resistance r,

00:04:23

which is given to us, and the total resistance of the

00:04:27

entire wire

00:04:28

is p 0, let’s set up the problem like this,

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determine this resistance, that is The

00:04:39

resistance of this piece is clear

00:04:43

that if these points are moved close, then

00:04:45

these two conductors will be connected to each

00:04:47

other and the resistance r will be small, 0,

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or very small,

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if we move these points apart, then

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this resistance will increase and

00:04:56

we need to

00:04:57

ensure that it is 80 By the way,

00:05:01

if these points are at opposite

00:05:03

ends of the ring, what will be the total

00:05:06

resistance then? 360 all together,

00:05:11

each half is 8 x 180, how are they

00:05:16

connected to each other?

00:05:24

get

00:05:27

a little less than 80, that is, these ends

00:05:30

need to be brought a little closer together and let’s

00:05:33

just denote this resistance with the letters r

00:05:35

large without a zero and try to find it,

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please tell me what is the resistance of this

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piece

00:05:51

without this and RF is large and what is the

00:05:56

resistance of this piece without this

00:05:59

this r zero minus r means that all of

00:06:04

us have a resistance

00:06:12

r zero minus

00:06:15

how this

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section and this section of the wire are connected to each other,

00:06:22

we have just discussed that they are

00:06:24

connected in parallel, therefore we

00:06:27

can

00:06:29

use the formula for

00:06:31

parallel connection of two different ones to find the total resistance

00:06:34

conductors, I will remind you of this formula, in

00:06:37

this corner we will write r the total is

00:06:40

equal to r 1 and r 2 divided by r1 + r2

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only instead of

00:06:50

r1 we will have this resistance and instead of

00:06:54

r2 this is then this formula

00:06:58

in relation to our problem we will write

00:07:00

so r is small and equals the fraction in the numerator

00:07:05

r multiplied by r 0 minus r in the denominator

00:07:16

r

00:07:19

add r 0 minus r

00:07:29

we connected two conductors in parallel with

00:07:33

resistance r and resistance r 0

00:07:37

minus r in the numerator of the product of these

00:07:40

two resistances denominator the sum as

00:07:43

you see here r cancels out,

00:07:46

let's now do this let's solve the equation that

00:07:49

here the unknown r is large, the

00:07:53

value r 0 is

00:07:55

known, the value r is known small, we need to find r

00:07:58

large let's proceed as

00:08:00

follows, multiply the left-right side by r

00:08:03

0

00:08:06

then we will disappear / we will get p

00:08:12

small by r 0 equals and here we will open the

00:08:16

brackets

00:08:18

p- p zero minus kr square let's look

00:08:25

at this equation through the eyes of mathematicians p

00:08:28

large unknown quantity

00:08:30

what we got what structure does

00:08:34

this equation have what would it be called

00:08:37

in mathematics

00:08:38

if instead of p there were x for example a

00:08:41

quadratic equation we get a

00:08:44

quadratic equation we come across this for the first time

00:08:46

in the course electricity, but

00:08:49

nevertheless, it is so, it means that

00:08:51

we need to write this quadratic

00:08:53

equation in a standard way

00:08:55

so that on the left there are all terms and on

00:08:57

the right there is 0, move the

00:09:00

r-square to the left, then move r 0

00:09:07

multiplied by r to the left, it will already be

00:09:10

minus minus r 0 multiplied by

00:09:15

r and we already have this free term here

00:09:18

+ r small by r 0 equals zero

00:09:23

with which we will compare this quadratic

00:09:26

equation with the usual one x square plus

00:09:31

bx plus c equals zero these are the reduced

00:09:36

quadratic equations

00:09:38

let's solve it using standard methods.

00:09:40

you remember the formula for the roots of a

00:09:42

quadratic equation like this x

00:09:48

equals minus b plus minus the root of b

00:09:54

square minus 4 father discriminant divide

00:09:58

by 2a we have unity in the role of

00:10:02

a here is the coefficient for the

00:10:09

unknown in the role of b we have

00:10:14

minus r 0 in the role of c we play p small r

00:10:22

zero

00:10:26

so let's now write this formula

00:10:29

for this equation, we get

00:10:33

r large

00:10:35

equals / in the denominator there will be 2 in

00:10:41

the numerator there will be r 0 plus minus the

00:10:47

square root of 0 square here we have

00:10:53

minus 4 share and the unit is ocelot 4 solution

00:11:02

zero

00:11:07

so

00:11:09

let's simplify this formula a little,

00:11:13

let's take out

00:11:15

from under the root p 0 the square equals

00:11:19

now you just need to be careful,

00:11:22

this is a two, I'll write it down as one second, it's

00:11:25

just r 0 here and here it is

00:11:30

plus or minus now I take out

00:11:33

p 0 square as the root it turns into

00:11:36

r zero

00:11:37

that under the root remains carefully

00:11:40

look at one minus

00:11:45

I put p 0 squared therefore

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and here only p 0 to the first power

00:11:50

will be therefore 4p small and p 0 will appear in

00:11:55

the denominator the bracket is closed now it

00:12:02

probably makes sense to put p 0 out of the

00:12:04

bracket and it turns out equals equals r

00:12:09

0 in half and in brackets

00:12:12

there will be such a beautiful form of

00:12:15

one plus minus instead of p zero

00:12:19

here are ones so we immediately write the root and

00:12:22

here there will be one minus 4p small by r

00:12:27

0 this is our working formula, we just have to

00:12:30

write it down completely

00:12:36

write r equals steering multiplied in half

00:12:44

on the bracket brackets one plus

00:12:50

minus the root of one minus 4p small

00:13:00

divide by r 0

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the bracket is closed in a frame

00:13:12

our quadratic equation has two

00:13:14

roots let's find both

00:13:17

and then think about how this happened

00:13:19

how this can be the problem is given a

00:13:22

clear answer 2

00:13:24

this can happen or not now

00:13:28

let's figure it out, so

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r equals r 0 in half 360 he divide by

00:13:40

2 multiplied by one plus minus the

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square root of units a minus 4 on the 80th

00:13:58

divided by r 0

00:14:01

360 he here he is her mind and will be reduced

00:14:06

to get a dimensionless value

00:14:08

close the bracket equals

00:14:12

360 in half to 180 oh and I’ll write it later and

00:14:18

here there will be one plus minus

00:14:21

well, let’s see what

00:14:22

happens here

00:14:25

320 divided by 360 32 36 them but how much

00:14:34

can you reduce

00:14:36

this 8 9 from 8 9 from one from one

00:14:43

subtract eight ninths how much is 1 9

00:14:46

it is under root of the root of 1 9 one third

00:14:49

plus minus one third therefore p 1 1

00:14:56

root of Ohm's equation

00:14:57

we add it will be 180

00:15:02

times 1 plus one third four

00:15:08

thirds

00:15:09

by four thirds he how much will it be

00:15:21

240p 2 equals 180

00:15:26

times what will be the parenthesis if

00:15:29

we take with minus two thirds two thirds

00:15:34

it turns out to be 120 he and so we can take

00:15:43

this piece with resistance you want

00:15:47

120 m you want 240 how it turns out how to

00:15:52

explain it Ariadne

00:15:55

[music]

00:16:04

good girl look guys if this is 120

00:16:08

m then this is 360 -120 how much it will be 240

00:16:14

here they are, that is, you can take here

00:16:17

120 then this piece will be 240 you can

00:16:20

take this piece 240 then the remaining

00:16:23

piece will be 120 so as you can see

00:16:26

mathematics describes this

00:16:30

problem perfectly both roots fit let's

00:16:35

move on from problems just for the

00:16:37

curious to the problems of the Olympiad

00:16:41

this is a Gelfgat problem book In it, the

00:16:45

Olympiad problems are marked with the letter q,

00:16:48

problem 40 from Gelfgat, number 40

00:16:58

Gelfgat,

00:17:18

what is the resistance of the circuit between points

00:17:21

a and b, the

00:17:22

resistance of each link is 2 ohms,

00:17:25

we mean such a diamond is

00:17:27

assembled from wires, each piece

00:17:30

of which has a resistance of 2 ohms,

00:17:33

let's denote this resistance with the

00:17:35

letter r is small and you need to find the total

00:17:42

resistance of the circuit between points a b b a

00:17:50

b let's

00:17:53

redraw the diagram

00:18:01

to the right of the figure, leave some free

00:18:04

space, first redraw this diagram in the

00:18:08

form in which it is given in the problem statement,

00:18:11

here's a diamond, its pieces are

00:18:15

divided here and the

00:18:16

sides are divided in half and connected

00:18:21

to each other like this and here too. there

00:18:28

are also connections here.

00:18:31

connected to an electrical circuit and

00:18:34

marked with us and this. connected to an

00:18:37

electrical circuit and marked with the letter b,

00:18:42

you need to find the resistance between points

00:18:45

a and b,

00:18:46

that is, what would a device measuring

00:18:49

resistance show, I already talked about it’s called a

00:18:51

meter, if we connected it here, this is the

00:18:54

task,

00:18:58

please tell me, can we say

00:19:01

that these are the conductors each of

00:19:04

the conductors can be depicted as a

00:19:06

resistor, but this is not necessarily true, can

00:19:08

we say that all these conductors are

00:19:10

connected in parallel, no, there are all

00:19:15

sorts of sections here, I don’t see here

00:19:18

actually a

00:19:21

pure-blooded parallel connection

00:19:23

or a pure-blooded serial connection, that

00:19:26

is, unlike those, it is impossible to ask what we

00:19:27

decided yesterday break this chain

00:19:31

into pieces

00:19:33

where some of the pieces are a

00:19:36

parallel connection and some are

00:19:38

in series, and in these cases

00:19:40

all sorts of interesting techniques are used,

00:19:43

one of them I’ll show you now, look,

00:19:47

let’s now move this point apart like

00:19:49

this, disconnect it and insert a

00:19:52

small wire here and redraw

00:19:57

this diagram

00:20:03

like this,

00:20:12

these points a and b

00:20:19

and that small wire

00:20:23

connects these two points like this,

00:20:25

that is, we broke the contact into

00:20:26

disconnected ones and inserted a thin wire into the break in the circuit,

00:20:30

and now

00:20:33

imagine that we connected point a to the

00:20:36

positive pole of the current source,

00:20:38

point b to the negative and let's trace in

00:20:42

which direction the currents flow in this circuit

00:20:45

along the wires

00:20:47

here the current will flow where here here here here

00:20:51

here

00:20:52

from the positive pole to

00:20:54

the negative here here here here here so it

00:21:02

will flow so it will flow and now the main

00:21:06

question is where will the current flow along this

00:21:08

wire ariana will it flow or it won’t

00:21:19

flow, it won’t flow, let’s say, how can you

00:21:24

prove it, let’s say it will flow to the left,

00:21:27

but excuse me, the diagram looks completely

00:21:30

symmetrical, if I

00:21:32

turn it over like this, it will already mean that the

00:21:34

current has flowed to the right, but the diagram

00:21:37

is combined with itself, which means it cannot

00:21:41

flow at the same time only to the left and

00:21:42

to the right, so there will be no current here, there will be

00:21:45

no current flowing here, they will

00:21:50

collide and everyone will go back

00:21:53

to their own countries with bruises, then it’s the

00:21:54

end

00:21:59

and now the main question is, if it’s

00:22:03

just not here, then when we remove this delay,

00:22:06

something will change, no, so this

00:22:10

delay can be thrown away and go to the

00:22:13

next diagram, here is a diamond and here there

00:22:22

is simply no electrical contact,

00:22:28

no contact. a point b and now

00:22:39

look what happened - a circuit

00:22:42

that can already be divided into parts into

00:22:47

this piece and this one are absolutely

00:22:52

identical and they are connected

00:22:54

in parallel to each other, the beginning of these two sections

00:22:58

connect points a and the ends to point b

00:23:01

since they are the same, then the total

00:23:03

resistance of the entire circuit will be two times

00:23:06

less than the resistance of this

00:23:08

section and now let's write the right

00:23:12

side, we won't consider it the

00:23:14

same as the left ones, let's write on each of the

00:23:17

pieces of wire

00:23:18

its electrical resistance, it is

00:23:20

designated by the letter r r r r and here

00:23:27

r

00:23:28

we look at this piece

00:23:32

like these two sections are connected to each other

00:23:35

in series

00:23:38

these two sections are in series what is the

00:23:42

resistance of the left piece 2r of the right

00:23:46

these two sections are

00:23:48

connected to each other in parallel know what the

00:23:51

total resistance r will be and so this whole

00:23:54

piece has a resistance r

00:24:00

this piece has a resistance r

00:24:03

it is connected to this With this piece,

00:24:05

what is the total resistance of

00:24:06

3p because here is a series

00:24:09

connection pprr, so these are all three r the

00:24:13

entire left side has a resistance of 3p

00:24:17

but the right side is no different,

00:24:21

it also has a resistance of 3p what

00:24:24

will be the total resistance

00:24:26

3p here 3p here how they are connected

00:24:31

in parallel at parallel connection of

00:24:33

identical ones is divided in half when

00:24:37

connecting two conductors in parallel, the

00:24:39

total resistance becomes half as

00:24:41

much and so we can write fish

00:24:45

equals 3 r in half three second

00:24:50

r this is the working formula we substitute fish

00:24:58

equals 3 divided by 2 multiplied by 2

00:25:03

ohms

00:25:05

equals 3rd a This is a technique that can be

00:25:11

used if you are calculating the

00:25:13

resistance of symmetrical circuits, that is,

00:25:16

you can break the pieces, turning the circuit

00:25:21

into a simpler one. Another problem is also the

00:25:26

Olympiads

00:25:28

44 44

00:25:33

and here are several problems, we won’t solve everything, we wo

00:25:35

n’t solve the problem about 4444.

00:26:00

Maxim listened to me attentively for a month

00:26:02

at the beginning lesson, I said that there

00:26:05

will be no such problems, what

00:26:12

about an Olympiad task, which means that if you

00:26:15

participate in the Olympiad,

00:26:18

you may come across such a task, but it won’t come up on the test,

00:26:21

this is for your development, not for a grade, we

00:26:25

guys are learning, but to understand what

00:26:28

is happening around and so problem number 44

00:26:33

and you see this cube

00:26:37

so you need to find the resistance of the

00:26:41

wire circle

00:26:42

between the points and here it is on the left and c1

00:26:46

on the right

00:26:47

let's pay attention that the

00:26:51

resistance of each edge on the 12th and so

00:26:54

p is small and on the

00:26:58

12th you need to find the resistance of the section

00:27:03

between points a and c 1 there the pen

00:27:08

blocked the unit

00:27:10

here and cd-r

00:27:15

ac1 draw a cube

00:27:40

everywhere here we put dots here there is an

00:27:44

electrical connection do not forget to

00:27:54

put dots on the test

00:27:56

where . and here we have it

00:28:00

connected to an electrical circuit. huh

00:28:04

. c1 is located here

00:28:09

c1 connected to the electrical circuit

00:28:12

you need to calculate the resistance of such a

00:28:17

wire cube if

00:28:21

you turn on a resistance meter between points a and c 1

00:28:25

here it is also impossible to identify a non-

00:28:29

parallel non-series

00:28:31

connection, however, we will solve this problem

00:28:33

now; for this we need to understand

00:28:38

what the nature of the symmetry is of this cube, is

00:28:44

it really a symmetrical cube, a

00:28:46

wonderful body, it has several axes of

00:28:50

symmetry, for example, 1

00:28:52

if I rotate it 90 degrees, it

00:28:55

will align itself, it's called a

00:28:59

fourth-order axis because 90

00:29:01

degrees is the fourth part of a circle,

00:29:02

but there is a less obvious axis of symmetry

00:29:07

if you look here so

00:29:10

please tell me how much I need to rotate

00:29:12

the cube so that it aligns with itself

00:29:15

by one third of a turn one third of a

00:29:21

turn means 120 degrees to

00:29:23

therefore if I rotate this cube

00:29:26

around this spatial

00:29:28

diagonal by 120 degrees then it

00:29:31

will align with itself and this is the

00:29:34

property cube, we are now using

00:29:36

this third-order axis, let's

00:29:39

do this: let's

00:29:40

run a current here, its strength will be and

00:29:45

it is clear that this current, having passed through the cube,

00:29:48

will come out through this conductor and here the

00:29:52

current strength will also be and in order to

00:29:56

create this current we will apply to cube

00:29:59

voltage which we denote by a letter and

00:30:02

apply how many volts it doesn’t matter and now

00:30:10

look here the current flows in and

00:30:14

this is the node and then the current branches out in

00:30:18

three directions please tell me the

00:30:21

current strength here here and here will be

00:30:23

different it will be the same due to the symmetry of the cube it

00:30:27

means it will go here by force and divide by

00:30:31

3

00:30:33

well the same current will go here and divide

00:30:39

by 3 the same current will go here and divide

00:30:42

by 3 we connect this this point to the

00:30:46

positive pole and this one to

00:30:48

the negative we went further these currents

00:30:52

flowing through the conductors of the cube then they will

00:30:56

come to this point they will come in how many

00:30:58

ways in three 1 2 3 means if

00:31:03

current flows here and then please tell me

00:31:07

what current will flow here

00:31:10

too and at 3

00:31:13

here current will flow to this node c1

00:31:18

too

00:31:19

and divide by 3 and current will also flow through this conductor

00:31:24

and at 3 go further

00:31:31

look at the

00:31:32

current and sodium in this node

00:31:36

is divided into two currents,

00:31:40

please tell me they will be different or

00:31:41

the

00:31:43

same, in this case, if there is current

00:31:46

and sodium, then what kind of current will flow through this

00:31:49

edge and divide by 6 and here the same amount,

00:31:53

but this is not of much interest to us, not

00:31:56

very interested

00:31:58

and now let's look at one of

00:32:01

the paths along which the current flowed from point

00:32:04

a of the night from the beginning to point c1 here it went

00:32:08

like this and on 3 here it went like this and on 6 here

00:32:13

it went like this and on 3 and apply Ohm's law to

00:32:17

each of these three sections we

00:32:20

will get next is . and this.

00:32:26

we have b let's put this point this is the point in

00:32:31

our churches how

00:32:35

these pieces are connected to each other in series the

00:32:38

voltage in this section the voltage in

00:32:41

this section and the voltage in this

00:32:43

section add up therefore the total

00:32:46

voltage is equal to the sum of and between a and b

00:32:52

plus the voltage between b and c and b c +

00:32:58

voltage - c between c&c one sc1 has

00:33:05

and now according to Ohm's law y a b is equal to the

00:33:10

current strength and by 3 times r and by 3

00:33:17

times the top on the section bc

00:33:21

the voltage is here the current is here and by

00:33:24

6 the resistance r is small plus and

00:33:28

multiplied by 6 by the resistance r is small

00:33:32

in the section cd1 current strength and multiplied by 3

00:33:40

by r is small we see that we can take it out of the

00:33:43

bracket and r oh of

00:33:51

course yes just y thank you yes I

00:33:56

already did step 2 right away I’m looking

00:33:58

at the clock I’m afraid not have time

00:33:59

so well, well, help further that there

00:34:03

will be one third + 1 6 plus one third

00:34:10

means 2 sixths + 1 6 plus 2 sixths

00:34:15

will be five sixths

00:34:16

equals five sixths r multiplied by and

00:34:24

so if we

00:34:28

apply a current here with force and then the voltage between

00:34:34

these two points according to Ohm's law will be like

00:34:36

this: how to find the total resistance r

00:34:40

or fac-1 we are looking for because we were looking for

00:34:46

it in the very first lesson when

00:34:47

we became acquainted with this quantity, the voltage

00:34:48

must be divided by the current strength y

00:34:53

divided by

00:34:55

and equals where to the neck here it is 5r divided

00:35:01

by 6 and and divided by and the current

00:35:06

will be reduced and we get r

00:35:12

ac1 equals five-sixths

00:35:16

thermal and here is our working formula that

00:35:21

remains to substitute the numbers r a c one

00:35:31

equals five-sixths

00:35:33

multiplied by 12 it equals 10 answer 10

00:35:42

it will now ring the bell

00:35:46

lesson is over

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