Download "Урок 253. Задачи на расчет электрических цепей - 3"

Please wait. We're preparing links for easy ad-free video watching and downloading.

Previous video: Урок 264. Соединение источников тока в батареи Next video: Урок 239. Соединение конденсаторов в батареи

Урок 239. Соединение конденсаторов в батареи

Урок 239. Соединение конденсаторов в батареи

Channel: Павел ВИКТОР

Урок 264. Соединение источников тока в батареи

Урок 264. Соединение источников тока в батареи

Channel: Павел ВИКТОР

Урок 192. Кипение жидкости

Урок 192. Кипение жидкости

Channel: Павел ВИКТОР

Урок 249. Последовательное и параллельное соединение проводников

Урок 249. Последовательное и параллельное соединение проводников

Channel: Павел ВИКТОР

Урок 250. Задачи на расчет электрических цепей - 1

Урок 250. Задачи на расчет электрических цепей - 1

Channel: Павел ВИКТОР

Урок 255. Задачи на работу и мощность электрического тока

Урок 255. Задачи на работу и мощность электрического тока

Channel: Павел ВИКТОР

Урок 160 (осн). Задачи на соединение проводников - 3

Урок 160 (осн). Задачи на соединение проводников - 3

Channel: Павел ВИКТОР

Урок 159 (осн). Задачи на соединение проводников - 2

Урок 159 (осн). Задачи на соединение проводников - 2

Channel: Павел ВИКТОР

Урок 181. Простейшие задачи на тепловые двигатели

Урок 181. Простейшие задачи на тепловые двигатели

Channel: Павел ВИКТОР

Электродвигатель постоянного тока. Принцип работы.

Электродвигатель постоянного тока. Принцип работы.

Channel: Halyk Smart

Закон Ома

Последовательное соединение

Параллельное соединение

Ришельевский лицей

00:00:10

today's two lessons we are built according to

00:00:13

this plan, now we are still solving problems on

00:00:16

calculating electrical circuits

00:00:18

in the next lesson I will tell you

00:00:21

something new and so the topic is

00:00:26

solving the problem,

00:00:37

homework, taking into account the

00:00:40

new material that will be in the second

00:00:42

lesson,

00:00:43

notes from Myakishev’s textbook, paragraph

00:00:55

106

00:01:00

Gelfgat problem book and one problem number 11 8 and

00:01:13

4 more problems according to Rymkevich with numbers 787 788

00:01:31

794 and 796

00:01:40

this is for tomorrow tomorrow we have 2

00:01:44

lessons in the second and fourth but

00:01:46

then we also have practice

00:01:48

for half the class and so we are working today

00:01:52

on the problem book gald farba let's start task

00:01:56

number 19 11 the task is accompanied by an

00:01:59

electrical circuit so let's

00:02:01

start with it and then write down a short

00:02:03

condition the circuit consists of three resistors

00:02:06

and this circuit is called a four-terminal circuit it

00:02:10

has two poles two inputs

00:02:17

and then there is some circuit and 2 outputs

00:02:29

here is the output inside the four-terminal circuit three

00:02:37

resistors their resistance r1 upper

00:02:51

r2 on the right side of the output

00:02:55

r3 and this is what is known about the properties of this

00:03:01

four-terminal network if a

00:03:04

voltage u1 of 100 volts is applied to the input of the electrical circuit, then the voltage

00:03:07

at the output of the op-amp is 340

00:03:16

volts if a voltage of 100 volts is applied to the input

00:03:23

u1 turn of resistor r1 then on the output

00:03:27

will be voltage u 3,

00:03:30

which is precisely the

00:03:31

voltage across resistor r3, such

00:03:34

designations we have, while

00:03:37

current i2 1 ampere flows through resistance r2,

00:03:45

this is the first piece of information, the

00:03:49

second piece of information, if

00:03:52

voltage u 3 is applied to the output of the circuit, stroke wipe

00:03:58

stroke 60 volts then the voltage at the input

00:04:05

turns out to be equal

00:04:07

to one stroke 15 volts, determine the

00:04:13

resistance values r1 r2 and r3

00:04:21

r1 r2 and r3 this is the task, so if

00:04:31

100 volts are applied to input a, the output will be

00:04:34

40, let’s imagine this better, let

00:04:38

’s draw the current source, it’s clear that this

00:04:42

cannot be a galvanic cell

00:04:43

just an ordinary galvanic cell

00:04:45

gives one and a half lithium 3 volts here

00:04:48

we definitely have a battery connected to the

00:04:53

input and it gives voltage u1 it

00:04:58

comes out while voltage u 3

00:05:01

can be measured with a voltmeter, of course, you

00:05:05

can of course not draw all this,

00:05:08

but it’s easier to imagine the condition of the

00:05:12

problem therefore it’s easier to solve,

00:05:13

also note that in such a connection, a

00:05:17

current i2 of 1 ampere flows through resistor r2,

00:05:21

let’s show it like this and now

00:05:27

let’s try, in this mode of

00:05:30

operation of the four-terminal network, to extract the maximum

00:05:32

information about the resistors that we know, we

00:05:38

know the voltage on this resistor is 100

00:05:42

volts and the voltage on this resistors

00:05:45

please tell me if we can then

00:05:47

find out the voltage on resistor r2 so

00:05:55

the blinker says that we need to subtract from u1 from

00:05:58

3 and we will get the voltage on the second

00:06:00

resistor let us designate it as u2 and

00:06:07

then really look at the voltage

00:06:10

here u2 the voltage here at 3

00:06:14

the voltmeter does not count this like a break in the

00:06:16

electrical color of an ideal voltmeter

00:06:18

while these two resistors are connected to each other

00:06:22

in series, in this case the

00:06:24

voltage between this point and this

00:06:26

voltage on the entire serial

00:06:28

chain is equal to the sum of y 2 + y 3 but this

00:06:32

voltage is exactly y 1 so we

00:06:35

find out that y 2 equals 1 minus y,

00:06:40

here’s the first step, but now it immediately

00:06:44

becomes clear how to find the

00:06:46

resistance r2 according to the conditions of the problem, the current

00:06:49

through this resistor is given, which means to

00:06:51

find the resistance r2, we’ll use

00:06:53

Ohm’s law, write out all three

00:06:56

variants of writing the classic Ohm’s law on the side

00:06:58

and divide y equally by r

00:07:01

further variations in the level and r and finally r is

00:07:07

equal to y divided at the bottom by the

00:07:11

voltage across the resistor r2 we know the strength of

00:07:14

the current we also know through it, which means we

00:07:17

can find the resistance r2

00:07:19

r2 equals q 2

00:07:23

ou-2 we have q1 minus 3 divided by

00:07:29

barely is

00:07:34

good we continue to think further, what

00:07:39

other ideas do you have? You can find r3

00:07:43

well, how do you divide the morning

00:07:49

into barely, and why do you divide the morning into barely a

00:07:56

series connection? with a

00:07:58

series connection, the currents through

00:08:00

both resistors will be the same, which means

00:08:04

here, too, the current dances,

00:08:06

turns here, yes, here, too, the current

00:08:08

will be barely,

00:08:10

therefore we know the voltage on this resistor,

00:08:13

by the condition a it is equal to three times the current

00:08:19

through this resistor, we also know by this

00:08:21

way we can say that p 3

00:08:25

is equal to the voltage on this resistor and

00:08:28

3 divided by barely

00:08:32

wounded 3 is ready to find the

00:08:39

resistance of the resistor r1 here it’s

00:08:42

not difficult guess that this circuit

00:08:44

will not be able to tell us anything about this

00:08:49

resistor because no matter what the voltage is

00:08:53

here, no matter what currents flow here,

00:08:56

this resistor is connected directly

00:08:58

to the current source, which means the voltage on

00:09:02

it is known about the strength of the current, we

00:09:05

cannot get information from this circuit,

00:09:07

we have already got everything we could, now all that remains is

00:09:10

to use the second part of the problem condition

00:09:13

when the four-terminal network is turned on

00:09:16

backwards, let's draw this circuit again, here is the

00:09:22

resistance r2, here are

00:09:26

the conclusions,

00:09:34

now the source is connected to the output,

00:09:42

here is the

00:09:46

source voltage, 3 of them have q3

00:09:51

stroke, let's finish drawing this resistor

00:09:55

though We won’t need it for work anymore,

00:10:00

this is resistor r2,

00:10:03

this is resistor r1, but this resistor r is

00:10:10

now you to the output, no, now this will be

00:10:13

the input, connect the voltmeter,

00:10:19

the voltmeter shows the voltage u1 stroke,

00:10:23

one stroke, let’s remember that

00:10:31

we already know r2, we want to find r1 what

00:10:35

ideas will we have,

00:10:41

let's climb along the same path

00:10:44

that we followed, what

00:10:45

we did with you, we found this

00:10:48

voltage as the difference, this is minus this,

00:10:51

let's denote this voltage with two

00:10:54

strokes and then this voltage u 3

00:11:05

stroke represents the sum of r2 y q2

00:11:09

stroke and one touch because, again,

00:11:12

this voltmeter does not show itself in any way,

00:11:15

that is, it does not branch off the current to itself,

00:11:17

so here

00:11:19

the same current flows through the resistor r1 and r2, and

00:11:22

since, among other things, they are

00:11:24

connected in series, the voltage

00:11:27

here plus the voltage here equals

00:11:31

voltage 3 stroke here you can do what you can do,

00:11:38

you can find the current strength through this

00:11:43

resistor marked with

00:11:45

two strokes and 2 strokes, the current strength through

00:11:54

this resistor

00:11:55

is equal to the voltage u2 stroke OU-2 stroke

00:11:59

this will be the difference of 3 pieces of them minus one

00:12:04

stroke u3 stroke minus y1 stroke divided

00:12:09

by resistance r2 like this and

00:12:15

resistance r2 we know, let's

00:12:18

take the value of r2 from here

00:12:20

then we will get it and two strokes

00:12:25

equals utrish 3 minus y1 stroke divided

00:12:32

by r 2 divided by this one let's write it like this y

00:12:34

one minus 3 will go here in

00:12:37

the denominator

00:12:41

but it will barely go into the numerator so that it turns out to be a

00:12:45

dimensionless fraction multiplied

00:12:47

by some value so and what does this

00:12:50

give us so we know the current through this

00:12:53

resistor

00:12:56

where it flows next, it flows all through

00:13:00

resistor r1 because the voltmeter does not

00:13:02

branch itself into the cake, which means there

00:13:05

will also be a current in force and 2 stroke voltage we

00:13:10

know the current strength we have already found it too, here it is the

00:13:15

current strength the voltage is given according to the conditions of

00:13:20

the problem, which means we can find

00:13:22

the resistance r1 we get from this

00:13:34

version of

00:13:35

Ohm’s law by dividing the voltage

00:13:37

voltage we have one stroke

00:13:41

one stroke per current strength through this

00:13:45

resistor then there are on and 2 pieces of them now all that

00:13:51

remains is just a stroke to take from here and we

00:13:55

get r1 equals the fraction and 2 stroke is

00:14:03

in our denominator so we turn the

00:14:06

fraction q1 minus u 3 divided by y 3 stroke minus

00:14:10

one stroke now this will barely go into

00:14:16

the denominator

00:14:20

and y is one stroke in the numerator here is 3, the formula is a

00:14:30

dimensionless fraction and here the ratio of

00:14:32

voltage to current has the dimension of

00:14:35

resistance, it remains to be calculated, well, let's

00:14:46

probably start with the card r3

00:14:50

equals the morning 40 volts divided by and 2 by

00:14:59

1 ampere equals 40 it is further r2r 2

00:15:10

equals the

00:15:12

fraction y 1 out of 100 minus 40 100 minus 40

00:15:23

divided by and 2 by 1 volt per ampere it

00:15:30

turns out 60 and finally r3

00:15:37

r3 equals 1 minus u 3 100 minus 40

00:15:45

100 minus 40 divided by y 3 prime minus

00:15:50

y1 prime 60 minus 15 units and I don’t write

00:15:57

they will already be reduced volts and with volts

00:16:00

one stroke 15 volts

00:16:05

divided by 1 ampere the answer will be in ohms

00:16:10

numerator 60 here 60 minutes 15-45 and here

00:16:15

we reduce 15 by 15 it turns out 60 divided

00:16:18

by 320

00:16:24

so resistance r1 here you need to

00:16:32

write r1 before

00:16:34

resistance r1 I use this formula

00:16:36

he considered this is 20 he resistance r2

00:16:41

this is 60 ohm resistance r 3 this is

00:16:45

40 and here is such a problem here we had the

00:16:49

opportunity to solve everything in a general form

00:16:51

because the lyrical tapeworm is especially

00:16:53

difficult if the circuit is complex it is possible to solve the

00:16:57

problem in parts it is not necessary to obtain

00:16:59

all the formulas in general,

00:17:01

now we will solve a similar

00:17:04

problem,

00:17:05

but on the contrary, we will not write a single formula,

00:17:08

that is, I will show you another style for

00:17:11

solving similar problems,

00:17:13

although on a test it is difficult to

00:17:15

convey your thoughts to the person who will

00:17:19

check it worked, but you can

00:17:22

quickly solve problems problem number 19 20

00:17:31

1920 farm

00:17:38

this electrical circuit is like this

00:17:44

and the values of the resistors are written directly on it

00:17:50

this resistor is 10 kohms

00:17:58

I won’t write units, we will all

00:18:02

mean in ohms then here’s another 10 kohms here

00:18:10

the resistor is also 10 m further the current source

00:18:24

is drawn here galvanic element, but

00:18:27

since it says seven and a half volts,

00:18:29

I’ll still write the batteries, I’ll draw a

00:18:32

battery of 7 and 5 volts, here the ammeter is turned on, here are the

00:18:46

connections, here are the connections, and finally, here’s

00:18:49

another resistor,

00:18:58

its resistance is 15 and now in

00:19:04

order for us to communicate we need

00:19:07

designations this resistor we will have r1

00:19:13

what could we call these elements

00:19:15

this r2

00:19:18

this r3hab what is below r4

00:19:26

well the question is the following: what

00:19:30

current flows through an ammeter

00:19:32

with a negligible internal

00:19:35

resistance

00:19:36

to the circuit shown shown in this

00:19:39

figure what is the current through the ammeter an

00:19:43

ideal ammeter with negligible

00:19:45

internal resistance

00:19:54

if you are dealing with an ammeter, you

00:19:57

can imagine that there is nothing here,

00:19:59

that is, there is just a wire going on and let's

00:20:03

try to find the distribution of currents in

00:20:06

this circuit, just if we replace

00:20:08

the ammeter with a conductor, it is easier to see

00:20:10

what type of connection is here and we will

00:20:15

look for currents not definitely right here

00:20:17

and in different parts of the circuit, look,

00:20:20

if it’s a vampire, there’s no meter, they forgot, and

00:20:24

resistor r1 is connected, its left

00:20:31

terminal is connected to the negative pole of

00:20:34

the battery, and the right terminal is just a

00:20:37

wire, and the right terminal is connected to

00:20:39

the positive, then this resistor is

00:20:41

simply connected to the current source

00:20:44

we know its resistance 5015 them we

00:20:48

know the voltage of the current source we can

00:20:50

find the current through this resistor it

00:20:55

does not depend on any other parts of

00:20:58

the circuit because the resistor is directly

00:21:00

connected to the current source

00:21:01

how to find the current here everything else

00:21:05

throw out the

00:21:08

voltage divide by the resistance

00:21:10

as you get half an ampere, let's

00:21:14

write the current flows from plus to minus,

00:21:17

which means here is the negative terminal, here

00:21:20

we have a current of 0.5 amperes, the

00:21:26

whole current is the same 0.5 amperes,

00:21:35

let's try to move forward with the project,

00:21:39

there is no ammeter, please tell me

00:21:42

how these two resistors are connected to each other,

00:21:45

what is it? -is it a standard connection or an

00:21:47

exotic one?

00:21:49

These two beginnings are connected to one

00:21:58

point in the electrical circuit and the two ends are

00:22:03

connected to another point in the electrical

00:22:06

circuit. How are they connected in parallel then

00:22:10

if each of them is 10 ohms then what is the

00:22:12

total resistance

00:22:14

5 means these two resistors together have a

00:22:18

resistance of 5th and now look

00:22:21

this pair of resistors is connected to

00:22:24

these resistors in what way

00:22:27

in series means these two

00:22:30

resistors look they are connected to the

00:22:32

positive pole then the current flows

00:22:35

through resistor r2 which is connected

00:22:38

in series

00:22:39

here 5 here 10 total resistance

00:22:43

15 what current flows through this resistor

00:22:48

also

00:22:49

that and a half divided by 15 half an ampere

00:22:55

05 amperes are

00:23:03

these half amperes

00:23:06

from which two identical

00:23:10

resistors are

00:23:11

connected in parallel, since they are connected in

00:23:14

parallel the voltage on them is the same, which

00:23:17

means the current strengths in them are also the same,

00:23:20

these two currents adding up give us half an

00:23:24

ampere each of these currents

00:23:26

what a fight for 025

00:23:30

so here a current of 025 amperes flows and here a

00:23:37

current of 025 amperes flows and they merge

00:23:43

to form a current of half an ampere,

00:23:45

well now we have come close to

00:23:49

solving the problem, look at this

00:23:52

node, a

00:23:56

current of half an ampere flows out of it along this branch and along this one

00:24:00

branches current is a quarter ampere then what

00:24:02

current flows through this conductor 075 the problem is

00:24:06

solved and equals 0 75 amperes here is a

00:24:12

completely different approach another style of

00:24:14

solving problems by the way you can

00:24:16

check yourself look let's come from the other

00:24:19

end we used this node

00:24:22

let's come from this end here

00:24:25

2 currents are combined half an ampere means through the

00:24:28

current source what kind of current flows

00:24:29

1 ampere 1 ampere now a quarter ampere leaves from this one

00:24:36

ampere to the left 025

00:24:43

amperes which means once from one ampere an

00:24:46

outflow of a quarter ampere branched off here

00:24:49

we have three quarters left of the

00:24:51

same 075 amperes 075

00:24:55

amperes it turns out and so you can

00:24:58

solve similar problems,

00:25:01

let's move on to another popular type of

00:25:12

problem for calculating the resistance of all sorts of

00:25:15

meshes problem number 19 16 number 1916

00:25:23

Hyde Park you need to find the resistance

00:25:32

between points a and b of a

00:25:34

wire mesh like this, draw a

00:25:40

rhombus like this, it is advisable that you

00:25:44

leave some space on the right because that now we

00:25:46

will redraw this rhombus and then

00:26:01

this whole connection is like this,

00:26:12

here the conductor comes off and here

00:26:17

the conductor comes off. and this is point b, the

00:26:22

resistance of each link of this grid

00:26:25

is small, you need to find the resistance of the entire

00:26:29

grid as a whole, how these elements of each of them are connected to each other,

00:26:37

these elements of each of them can

00:26:40

be considered a resistor, points must be

00:26:42

placed everywhere, that is, here is a resistor that is

00:26:46

simply not drawn here, these are all

00:26:48

resistors, each has the resistance r is

00:26:50

small

00:26:51

as they are connected to each other in

00:26:53

parallel in series in different ways and

00:26:57

not in parallel and inconsistently,

00:26:59

so if we want to solve this problem

00:27:02

we need to use some kind of trick, we

00:27:04

need to somehow redraw

00:27:08

this circuit, somehow modify it, but in

00:27:10

such a way that the resistance and at the same time, it did

00:27:13

n’t change, you can do the following,

00:27:17

imagine that you cut

00:27:21

here and here

00:27:23

this diagram was torn apart like this, let’s

00:27:28

see what happens, redraw

00:27:31

the diamond again

00:27:38

and now here we have a gap, so

00:27:50

and so,

00:27:54

and here we have a gap,

00:28:08

everything else remains the same -as before. b.

00:28:17

and here is the connection and now

00:28:31

please tell me is there a potential difference

00:28:35

between these two points here on

00:28:38

these streets oh no the circuit is symmetrical

00:28:41

so the voltage between this point and

00:28:44

this this this is the

00:28:47

same imagine that we

00:28:49

connected with a wire here in which

00:28:52

direction the current will flow to the left or to the right

00:28:55

considerations of symmetry will not flow anywhere,

00:28:58

therefore, if we connect with

00:29:01

a wire, that is, if we restore the

00:29:03

previous configuration, then the electrical

00:29:06

properties will remain the same, these

00:29:08

points we will call these

00:29:10

potential these potential points

00:29:13

can be connected

00:29:14

or can be broken, so this circuit will

00:29:19

have exactly the same resistance is like

00:29:21

this, but this is a simpler diagram,

00:29:27

now let’s use this same idea further

00:29:29

where else we can make a break here

00:29:38

and draw this diamond again with

00:29:44

these additional breaks we’ll

00:29:46

get

00:29:55

. b. and then this section

00:30:06

now looks like this, this section looks

00:30:12

now like this, here we have a gap, which means this is

00:30:19

what this section will look like, so and

00:30:23

so, and this section here,

00:30:30

we draw the gap like this, so and so,

00:30:43

we make electrical connections,

00:30:55

the resistance of this circuit will be the same

00:30:58

because here, too, there is no

00:31:01

potential difference for reasons of

00:31:03

symmetry,

00:31:04

that is, if we connect these two points, it

00:31:06

won’t flow like that, so it doesn’t matter whether there are

00:31:08

connections here or not, it was here,

00:31:12

and so the resistance of this chain and this

00:31:15

chain is the same, each link has a

00:31:17

resistance r small in addition, pay

00:31:21

attention that this whole chain consists of two

00:31:24

halves on the left and exactly the same on the right, which means that

00:31:27

if we find the resistance of the left

00:31:29

half, then the resistance of the right half will be

00:31:31

exactly the same, and so let’s deal with the left

00:31:33

half, what does it consist of, the

00:31:36

resistance of this piece r

00:31:39

the resistance of this piece era how

00:31:42

they are connected to each other

00:31:44

in series what is the total

00:31:46

resistance 2r to these two r

00:31:51

but exactly the same

00:31:53

resistance 2r how are they connected to each other

00:31:56

in parallel 2r connected

00:32:00

in parallel to 2r how much will be r and so this

00:32:04

whole little diamond between

00:32:07

these points

00:32:08

has a resistance r now look at

00:32:11

this diamond, this link is

00:32:14

connected in parallel or in

00:32:15

series, and this

00:32:18

link, this diamond has a resistance r,

00:32:21

each of these links also has a

00:32:24

resistance r,

00:32:25

they are connected in series, which means

00:32:28

together, this entire section of the circuit, as I see it in

00:32:31

dotted lines,

00:32:35

has a resistance of

00:32:38

3p here this is r plus this is r and this is r3 r3

00:32:49

we take the next step here is this piece of

00:32:54

wire what resistance does it have 4 1

00:32:59

2 3 4 sections power means here we have 4

00:33:03

r and this section is connected to 3p

00:33:07

as in parallel means total

00:33:10

resistance let's take a different

00:33:12

color chalk total resistance this

00:33:14

section

00:33:23

is the resistance of two resistors, one

00:33:26

of which is 4r, this second one is 3r, which means

00:33:29

we get 4 r multiplied by 3 r

00:33:35

divided by 4 r plus r numerator 12 in the

00:33:43

denominator 7 equals 12 sevenths r this is the

00:33:49

resistance between these two

00:33:51

points 12 sevenths we go further from this

00:33:59

point to point b there is a section with

00:34:02

resistance r and here a section with

00:34:05

resistance r

00:34:06

means 12 sevenths we still need to add 2r

00:34:10

and we get 12 sevenths r + 2 r

00:34:17

equals how much 2r is 14 7 and 12

00:34:24

sevenths twenty-six sevenths

00:34:26

twenty-six sevenths is what this is the

00:34:32

resistance of this left half of the

00:34:35

electrical circuit,

00:34:36

but exactly the same resistance is on the

00:34:39

right half and they are connected to

00:34:41

each other in parallel then what will be the

00:34:44

total resistance to p and twenty-

00:34:53

six-sevenths in half, that is, 13 sevenths

00:34:57

equals 1 2 multiplied by I’m just

00:35:01

writing this for clarity twenty-six

00:35:04

sevenths of r or r & b equals 13 sevenths of

00:35:14

r little

00:35:17

so what idea did we use now and

00:35:20

cry potential points can be connected

00:35:23

or breaks can be made cuts in an

00:35:27

electrical circuit to form

00:35:29

these potential points no current flows through them

00:35:32

therefore their

00:35:35

presence of connections there or the lack of

00:35:37

connections does not in any way affect the overall

00:35:39

resistance of the chain, well, let's

00:35:41

take a break and in the next lesson

00:35:43

we will deal with new material

Description:

Урок физики в Ришельевском лицее

Preparing download options

Popular

HD video

Only sound

All

* — If the video is playing in a new tab, go to it, then right-click on the video and select "Save video as..."

** — Link intended for online playback in specialized players

Questions about downloading video

How can I download "Урок 253. Задачи на расчет электрических цепей - 3" video?

http://unidownloader.com/ website is the best way to download a video or a separate audio track if you want to do without installing programs and extensions.
The UDL Helper extension is a convenient button that is seamlessly integrated into YouTube, Instagram and OK.ru sites for fast content download.
UDL Client program (for Windows) is the most powerful solution that supports more than 900 websites, social networks and video hosting sites, as well as any video quality that is available in the source.
UDL Lite is a really convenient way to access a website from your mobile device. With its help, you can easily download videos directly to your smartphone.

Which format of "Урок 253. Задачи на расчет электрических цепей - 3" video should I choose?

The best quality formats are FullHD (1080p), 2K (1440p), 4K (2160p) and 8K (4320p). The higher the resolution of your screen, the higher the video quality should be. However, there are other factors to consider: download speed, amount of free space, and device performance during playback.

Why does my computer freeze when loading a "Урок 253. Задачи на расчет электрических цепей - 3" video?

The browser/computer should not freeze completely! If this happens, please report it with a link to the video. Sometimes videos cannot be downloaded directly in a suitable format, so we have added the ability to convert the file to the desired format. In some cases, this process may actively use computer resources.

How can I download "Урок 253. Задачи на расчет электрических цепей - 3" video to my phone?

You can download a video to your smartphone using the website or the PWA application UDL Lite. It is also possible to send a download link via QR code using the UDL Helper extension.

How can I download an audio track (music) to MP3 "Урок 253. Задачи на расчет электрических цепей - 3"?

The most convenient way is to use the UDL Client program, which supports converting video to MP3 format. In some cases, MP3 can also be downloaded through the UDL Helper extension.

How can I save a frame from a video "Урок 253. Задачи на расчет электрических цепей - 3"?

This feature is available in the UDL Helper extension. Make sure that "Show the video snapshot button" is checked in the settings. A camera icon should appear in the lower right corner of the player to the left of the "Settings" icon. When you click on it, the current frame from the video will be saved to your computer in JPEG format.

What's the price of all this stuff?

It costs nothing. Our services are absolutely free for all users. There are no PRO subscriptions, no restrictions on the number or maximum length of downloaded videos.