In 2018, in the KIMs of the Unified State Examination in Physics, students will again find 32 tasks. Recall that in 2017 the number of tasks was reduced to 31. An additional task will be a question on astronomy, which, by the way, is again being introduced as a compulsory subject. It is not entirely clear, however, due to what hours, but, most likely, physics will suffer. So, if in the 11th grade you do not count the lessons, then the ancient science of the stars is probably to blame. Accordingly, you will have to prepare more on your own, because the volume of school physics will be extremely small in order to somehow pass the exam. But let's not talk about sad things.

The question on astronomy is number 24 and the first test part ends with it. The second part, respectively, has shifted and now begins with the 25th issue. Other than that, no major changes were found. The same short answer questions, matching and multiple choice tasks, and, of course, short and long answer tasks.

Exam tasks cover the following sections of physics:

1. Mechanics(kinematics, dynamics, statics, conservation laws in mechanics, mechanical oscillations and waves).

Molecular physics(molecular-kinetic theory, thermodynamics).

Electrodynamics and fundamentals of SRT(electric field, direct current, magnetic field, electromagnetic induction, electromagnetic oscillations and waves, optics, fundamentals of SRT).

The quantum physics(particle-wave dualism, physics of the atom and atomic nucleus).

2. Elements of astrophysics(solar system, stars, galaxies and universe)

Below you can familiarize yourself with the approximate tasks of the USE in 2018 in a demo version from FIPI. As well as familiarize yourself with the codifier and specification.

In 2018, graduates of grade 11 and institutions of secondary vocational education will take the USE 2018 in physics. The latest news regarding the Unified State Examination in Physics in 2018 is based on the fact that some changes will be made to it, both major and minor.

What is the meaning of changes and how many of them

The main change related to the Unified State Examination in Physics, compared to previous years, is the absence of a test part with a choice of answers. This means that preparation for the exam should be accompanied by the student's ability to give short or detailed answers. Therefore, it will no longer be possible to guess the option and score a certain number of points and you will have to work hard.

A new task 24 has been added to the basic part of the exam in physics, which requires the ability to solve problems in astrophysics. By adding No. 24, the maximum primary score has increased to 52. The exam is divided into two parts according to difficulty levels: a basic one of 27 tasks, involving a short or full answer. In the second part there are 5 tasks of an advanced level, where you need to give a detailed answer and explain the course of your solution. One important nuance: many students skip this part, but even trying to complete these tasks can get from one to two points.

All changes in the exam in physics are made in order to deepen the preparation and improve the assimilation of knowledge in the subject. In addition, the elimination of the test part motivates future applicants to accumulate knowledge more intensively and reason logically.

Exam Structure

Compared to the previous year, the structure of the USE has not changed significantly. 235 minutes are allotted for the entire work. Each task of the basic part should be solved from 1 to 5 minutes. Tasks of increased complexity are solved in about 5-10 minutes.

All CIMs are stored at the exam venue and will be opened during the test. The structure is as follows: 27 basic tasks test the knowledge of the examinee in all areas of physics, from mechanics to quantum and nuclear physics. In 5 tasks of a high level of complexity, the student shows skills in the logical justification of his decision and the correctness of the train of thought. The number of primary points can reach a maximum of 52. Then they are recalculated within the framework of a 100-point scale. Due to the change in the primary score, the minimum passing score may also change.

Demo version

The demo version of the exam in physics is already on the official fipi portal, which is developing a unified state exam. The structure and complexity of the demo version is similar to the one that will appear on the exam. Each task is described in detail, at the end there is a list of answers to questions on which the student checks his decisions. Also at the end is a detailed layout for each of the five tasks, indicating the number of points for correctly or partially completed actions. For each task of high complexity, you can get from 2 to 4 points, depending on the requirements and the deployment of the solution. Tasks can contain a sequence of numbers that you need to write down correctly, establishing a correspondence between elements, as well as small tasks in one or two actions.

• Download demo: ege-2018-fiz-demo.pdf
• Download archive with specification and coding: ege-2018-fiz-demo.zip

We wish you to successfully pass physics and enter the desired university, everything is in your hands!

Secondary general education

Getting ready for the Unified State Exam-2018: analysis of the demo version in physics

USE-2018. Physics. Thematic training tasks

The edition contains:
tasks of different types on all topics of the exam;
answers to all questions.
The book will be useful both for teachers: it makes it possible to effectively organize the preparation of students for the exam directly in the classroom, in the process of studying all topics, and for students: training tasks will allow you to systematically, when passing each topic, prepare for the exam.

A point body at rest begins to move along the axis Ox. The figure shows a projection dependency graph ax acceleration of this body with time t.

Determine the distance traveled by the body in the third second of motion.

Answer: _________ m.

Solution

Being able to read graphs is very important for every student. The question in the problem is that it is required to determine from the graph the dependence of the projection of acceleration on time, the path that the body has traveled in the third second of motion. The graph shows that in the time interval from t 1 = 2 s to t 2 = 4 s, acceleration projection is zero. Consequently, the projection of the resultant force in this area, according to Newton's second law, is also equal to zero. We determine the nature of the movement in this area: the body moved uniformly. The path is easy to determine, knowing the speed and time of movement. However, in the interval from 0 to 2 s, the body moved uniformly accelerated. Using the definition of acceleration, we write the velocity projection equation Vx = V 0x + a x t; since the body was initially at rest, then the velocity projection by the end of the second second became

Then the path traveled by the body in the third second

Answer: 8 m

Rice. 1

On a smooth horizontal surface lie two bars connected by a light spring. To a bar of mass m= 2 kg apply a constant force equal in modulus F= 10 N and directed horizontally along the axis of the spring (see figure). Determine the modulus of the elastic force of the spring at the moment when this bar moves with an acceleration of 1 m / s 2.

Answer: _________ N.

Solution

Horizontally on a body of mass m\u003d 2 kg, two forces act, this is the force F= 10 N and elastic force, from the side of the spring. The resultant of these forces imparts acceleration to the body. We choose a coordinate line and direct it along the action of the force F. Let's write down Newton's second law for this body.

Projected onto axis 0 X: F – F extr = ma (2)

We express from formula (2) the modulus of the elastic force F extr = F – ma (3)

Substitute the numerical values ​​into formula (3) and get, F control \u003d 10 N - 2 kg 1 m / s 2 \u003d 8 N.

Answer: 8 N.

Task 3

A body with a mass of 4 kg, located on a rough horizontal plane, was reported along it with a speed of 10 m / s. Determine the modulus of work done by the friction force from the moment the body begins to move until the moment when the speed of the body decreases by 2 times.

Answer: _________ J.

Solution

The force of gravity acts on the body, the reaction force of the support is the friction force that creates a braking acceleration. The body was initially reported with a speed equal to 10 m / s. Let's write down Newton's second law for our case.

Equation (1) taking into account the projection on the selected axis Y will look like:

N – mg = 0; N = mg (2)

In the projection on the axis X: –F tr = - ma; F tr = ma; (3) We need to determine the modulus of work of the friction force by the time when the speed becomes half as much, i.e. 5 m/s. Let's write a formula for calculating work.

A · ( F tr) = – F tr S (4)

To determine the distance traveled, we take the timeless formula:

Substitute (3) and (5) into (4)

Then the modulus of work of the friction force will be equal to:

Let's substitute numerical values

Answer: 150 J

USE-2018. Physics. 30 practice exam papers

The edition contains:
30 training options for the exam
instructions for implementation and evaluation criteria
answers to all questions
Training options will help the teacher to organize preparation for the exam, and students to independently test their knowledge and readiness for the final exam.

The stepped block has an outer pulley with a radius of 24 cm. Weights are suspended from the threads wound on the outer and inner pulleys as shown in the figure. There is no friction in the axis of the block. What is the radius of the inner pulley of the block if the system is in equilibrium?

Rice. one

Answer: _________ see

Solution

According to the condition of the problem, the system is in equilibrium. On the image L 1 , shoulder strength L 2 shoulder of force Balance condition: the moments of forces rotating the bodies clockwise must be equal to the moments of forces rotating the body counterclockwise. Recall that the moment of force is the product of the modulus of force and the arm. The forces acting on the threads from the side of the loads differ by a factor of 3. This means that the radius of the inner pulley of the block differs from the outer one also by 3 times. Therefore, the shoulder L 2 will be equal to 8 cm.

Answer: 8 cm

Task 5

Oh, at different times.

Select from the list below two correct statements and indicate their numbers.

1. The potential energy of the spring at time 1.0 s is maximum.
2. The period of oscillation of the ball is 4.0 s.
3. The kinetic energy of the ball at time 2.0 s is minimal.
4. The amplitude of the ball oscillations is 30 mm.
5. The total mechanical energy of the pendulum, consisting of a ball and a spring, is at a minimum at 3.0 s.

Solution

The table shows data on the position of a ball attached to a spring and oscillating along a horizontal axis. Oh, at different times. We need to analyze this data and choose the right two statements. The system is a spring pendulum. At the point in time t\u003d 1 s, the displacement of the body from the equilibrium position is maximum, which means this is the amplitude value. by definition, the potential energy of an elastically deformed body can be calculated by the formula

where k- coefficient of spring stiffness, X- displacement of the body from the equilibrium position. If the displacement is maximum, then the speed at this point is zero, which means that the kinetic energy will be zero. According to the law of conservation and transformation of energy, potential energy should be maximum. From the table we see that the body passes half of the oscillation for t= 2 s, total oscillation in twice the time T= 4 s. Therefore statements 1 will be true; 2.

Task 6

A small piece of ice was lowered into a cylindrical glass of water to float. After some time, the ice completely melted. Determine how the pressure on the bottom of the glass and the water level in the glass have changed as a result of the melting of the ice.

1. increased;
2. decreased;
3. hasn't changed.

Write to table

Solution

Rice. one

Problems of this type are quite common in different versions of the exam. And as practice shows, students often make mistakes. Let's try to analyze this task in detail. Denote m is the mass of a piece of ice, ρ l is the density of ice, ρ w is the density of water, V pt is the volume of the immersed part of the ice, equal to the volume of the displaced liquid (volume of the hole). Mentally remove the ice from the water. Then a hole will remain in the water, the volume of which is equal to V pm, i.e. volume of water displaced by a piece of ice one( b).

Let us write down the condition of ice floating Fig. one( a).

Fa = mg (1)

ρ in V pm g = mg (2)

Comparing formulas (3) and (4) we see that the volume of the hole is exactly equal to the volume of water obtained from the melting of our piece of ice. Therefore, if we now (mentally) pour the water obtained from ice into the hole, then the hole will be completely filled with water, and the water level in the vessel will not change. If the water level does not change, then the hydrostatic pressure (5), which in this case depends only on the height of the liquid, will also not change. Therefore, the answer will be

USE-2018. Physics. Training tasks

The publication is addressed to high school students to prepare for the exam in physics.
The allowance includes:
20 training options
answers to all questions
USE answer forms for each option.
The publication will assist teachers in preparing students for the exam in physics.

A weightless spring is located on a smooth horizontal surface and is attached to the wall at one end (see figure). At some point in time, the spring begins to deform, applying an external force to its free end A and uniformly moving point A.

Establish a correspondence between the graphs of dependences of physical quantities on deformation x springs and these values. For each position in the first column, select the corresponding position from the second column and write in table

Solution

It can be seen from the figure for the problem that when the spring is not deformed, then its free end, and accordingly point A, are in a position with the coordinate X 0 . At some point in time, the spring begins to deform, applying an external force to its free end A. Point A moves uniformly. Depending on whether the spring is stretched or compressed, the direction and magnitude of the elastic force arising in the spring will change. Accordingly, under the letter A), the graph is the dependence of the elastic modulus on the deformation of the spring.

The graph under the letter B) is the dependence of the projection of the external force on the magnitude of the deformation. Because with an increase in the external force, the magnitude of the deformation and the elastic force increase.

Answer: 24.

Task 8

When constructing the Réaumur temperature scale, it is assumed that at normal atmospheric pressure, ice melts at a temperature of 0 degrees Réaumur (°R), and water boils at a temperature of 80°R. Find the average kinetic energy of the translational thermal motion of an ideal gas particle at a temperature of 29°R. Express your answer in eV and round to the nearest hundredth.

Answer: _______ eV.

Solution

The problem is interesting in that it is necessary to compare two temperature measurement scales. These are the Réaumur temperature scale and the Celsius temperature scale. The melting points of ice are the same on the scales, but the boiling points are different, we can get a formula for converting degrees Réaumur to degrees Celsius. it

Let's convert the temperature of 29 (°R) to degrees Celsius

We translate the result into Kelvin using the formula

T = t°C + 273 (2);

T= 36.25 + 273 = 309.25 (K)

To calculate the average kinetic energy of the translational thermal motion of particles of an ideal gas, we use the formula

where k– Boltzmann constant equal to 1.38 10 –23 J/K, T is the absolute temperature on the Kelvin scale. It can be seen from the formula that the dependence of the average kinetic energy on temperature is direct, that is, how many times the temperature changes, the average kinetic energy of the thermal motion of molecules changes so many times. Substitute the numerical values:

The result is converted to electron volts and rounded to the nearest hundredth. Let's remember that

1 eV \u003d 1.6 10 -19 J.

For this

Answer: 0.04 eV.

One mole of a monatomic ideal gas is involved in process 1–2, the graph of which is shown in VT-diagram. Determine for this process the ratio of the change in the internal energy of the gas to the amount of heat imparted to the gas.

Answer: ___________ .

Solution

According to the condition of the problem in process 1–2, the graph of which is shown in VT-diagram, one mole of a monatomic ideal gas is involved. To answer the question of the problem, it is necessary to obtain expressions for changing the internal energy and the amount of heat imparted to the gas. Isobaric process (Gay-Lussac law). The change in internal energy can be written in two forms:

For the amount of heat imparted to the gas, we write the first law of thermodynamics:

Q 12 = A 12+∆ U 12 (5),

where A 12 - gas work during expansion. By definition, work is

A 12 = P 0 2 V 0 (6).

Then the amount of heat will be equal, taking into account (4) and (6).

Q 12 = P 0 2 V 0 + 3P 0 · V 0 = 5P 0 · V 0 (7)

Let's write the relation:

Answer: 0,6.

The reference book contains in full the theoretical material on the course of physics, which is necessary for passing the exam. The structure of the book corresponds to the modern codifier of content elements in the subject, on the basis of which the examination tasks are compiled - control and measuring materials (CMM) of the Unified State Examination. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks corresponding to the USE format. This will help the teacher to organize preparation for the unified state exam, and students to independently test their knowledge and readiness for the final exam.

A blacksmith forges an iron horseshoe weighing 500 g at a temperature of 1000°C. Having finished forging, he throws the horseshoe into a vessel of water. There is a hiss, and steam rises from the vessel. Find the mass of water that evaporates when a hot horseshoe is immersed in it. Consider that the water is already heated to boiling point.

Answer: _________

Solution

To solve the problem, it is important to remember the heat balance equation. If there are no losses, then heat transfer of energy occurs in the system of bodies. As a result, the water evaporates. Initially, the water was at a temperature of 100 ° C, which means that after the hot horseshoe is immersed, the energy received by the water will go immediately to vaporization. We write the heat balance equation

With and · m P · ( t n - 100) = lm in 1),

where L is the specific heat of vaporization, m c is the mass of water that has turned into steam, m p is the mass of the iron horseshoe, With g is the specific heat capacity of iron. From formula (1) we express the mass of water

When recording the answer, pay attention to what units you want to leave the mass of water.

Answer: 90

One mole of a monatomic ideal gas is involved in a cyclic process, the graph of which is shown in TV- chart.

Select two correct statements based on the analysis of the presented graph.

1. Gas pressure in state 2 is greater than gas pressure in state 4
2. The gas work in section 2–3 is positive.
3. In section 1–2, the gas pressure increases.
4. In section 4–1, a certain amount of heat is removed from the gas.
5. The change in the internal energy of the gas in section 1–2 is less than the change in the internal energy of the gas in section 2–3.

Solution

This type of task tests the ability to read graphs and explain the presented dependence of physical quantities. It is important to remember how dependency graphs look for isoprocesses in different axes, in particular R= const. In our example on TV The diagram shows two isobars. Let's see how the pressure and volume will change at a fixed temperature. For example, for points 1 and 4 lying on two isobars. P 1 . V 1 = P 4 . V 4 , we see that V 4 > V 1 means P 1 > P four . State 2 corresponds to pressure P one . Consequently, the gas pressure in state 2 is greater than the gas pressure in state 4. In section 2–3, the process is isochoric, the gas does no work, it is equal to zero. The assertion is incorrect. In section 1-2, the pressure increases, also incorrect. Just above we showed that this is an isobaric transition. In section 4–1, a certain amount of heat is removed from the gas in order to maintain the temperature constant when the gas is compressed.

Answer: 14.

The heat engine works according to the Carnot cycle. The temperature of the refrigerator of the heat engine was increased, leaving the temperature of the heater the same. The amount of heat received by the gas from the heater per cycle has not changed. How did the efficiency of the heat engine and the work of the gas per cycle change?

For each value, determine the appropriate nature of the change:

1. increased
2. decreased
3. hasn't changed

Write to table selected figures for each physical quantity. Numbers in the answer may be repeated.

Solution

Heat engines operating on the Carnot cycle are often found in assignments on the exam. First of all, you need to remember the formula for calculating the efficiency factor. Be able to record it through the temperature of the heater and the temperature of the refrigerator

in addition to be able to write the efficiency through the useful work of the gas A g and the amount of heat received from the heater Q n.

We carefully read the condition and determined which parameters were changed: in our case, we increased the temperature of the refrigerator, leaving the temperature of the heater the same. Analyzing formula (1), we conclude that the numerator of the fraction decreases, the denominator does not change, therefore, the efficiency of the heat engine decreases. If we work with formula (2), we will immediately answer the second question of the problem. The work of the gas per cycle will also decrease, with all the current changes in the parameters of the heat engine.

Answer: 22.

negative charge - qQ and negative- Q(see picture). Where is it directed relative to the picture ( right, left, up, down, towards the observer, away from the observer) charge acceleration - q in this moment of time, if only charges act on it + Q and –Q? Write your answer in word(s)

Solution

Rice. one

negative charge - q is in the field of two fixed charges: positive + Q and negative- Q, as shown in the figure. in order to answer the question of where the acceleration of the charge is directed - q, at the moment when only +Q and - charges act on it Q it is necessary to find the direction of the resulting force, as a geometric sum of forces According to Newton's second law, it is known that the direction of the acceleration vector coincides with the direction of the resulting force. The figure shows a geometric construction to determine the sum of two vectors. The question arises why the forces are directed in this way? Recall how similarly charged bodies interact, they repel each other, the Coulomb force of the interaction of charges is the central force. the force with which oppositely charged bodies attract. From the figure, we see that the charge is q equidistant from fixed charges whose moduli are equal. Therefore, modulo will also be equal. The resulting force will be directed relative to the figure way down. Charge acceleration will also be directed - q, i.e. way down.

Answer: Way down.

The book contains materials for the successful passing of the exam in physics: brief theoretical information on all topics, tasks of different types and levels of complexity, solving problems of an increased level of complexity, answers and evaluation criteria. Students do not have to search for additional information on the Internet and buy other manuals. In this book, they will find everything they need to independently and effectively prepare for the exam. The publication contains tasks of different types on all topics tested at the exam in physics, as well as solving problems of an increased level of complexity. The publication will provide invaluable assistance to students in preparing for the exam in physics, and can also be used by teachers in organizing the educational process.

Two resistors connected in series with a resistance of 4 ohms and 8 ohms are connected to a battery, the voltage at the terminals of which is 24 V. What thermal power is released in a resistor of a smaller rating?

Answer: _________ Tue.

Solution

To solve the problem, it is desirable to draw a series connection diagram of resistors. Then remember the laws of the series connection of conductors.

The scheme will be as follows:

Where R 1 = 4 ohm, R 2 = 8 ohms. The voltage at the battery terminals is 24 V. When the conductors are connected in series, the current strength will be the same in each section of the circuit. The total resistance is defined as the sum of the resistances of all resistors. According to Ohm's law for the circuit section we have:

To determine the thermal power released on a resistor of a smaller rating, we write:

P = I 2 R\u003d (2 A) 2 4 Ohm \u003d 16 W.

Answer: P= 16 W.

A wire frame with an area of ​​2 · 10–3 m 2 rotates in a uniform magnetic field around an axis perpendicular to the magnetic induction vector. The magnetic flux penetrating the frame area changes according to the law

Ф = 4 10 –6 cos10π t,

where all quantities are expressed in SI. What is the modulus of magnetic induction?

Answer: ________________ mT.

Solution

The magnetic flux changes according to the law

Ф = 4 10 –6 cos10π t,

where all quantities are expressed in SI. You need to understand what magnetic flux is in general and how this value is related to the magnetic induction modulus B and frame area S. Let's write the equation in general form to understand what quantities are included in it.

Φ = Φ m cosω t(1)

Remember that before the cos or sin sign there is an amplitude value of a changing value, which means Φ max \u003d 4 10 -6 Wb, on the other hand, the magnetic flux is equal to the product of the magnetic induction modulus and the circuit area and the cosine of the angle between the normal to the circuit and the magnetic induction vector Φ m = AT · S cosα, the flux is maximum at cosα = 1; express the modulus of induction

The answer must be written in mT. Our result is 2 mT.

Answer: 2.

The section of the electrical circuit is a series-connected silver and aluminum wires. A constant electric current of 2 A flows through them. The graph shows how the potential φ changes in this section of the circuit when it is displaced along the wires by a distance x

Using the graph, select two correct statements and indicate their numbers in the answer.

1. The cross-sectional areas of the wires are the same.
2. Cross-sectional area of ​​silver wire 6.4 10 -2 mm 2
3. Cross-sectional area of ​​silver wire 4.27 10 -2 mm 2
4. A thermal power of 2 W is released in the aluminum wire.
5. Silver wire produces less thermal power than aluminum wire.

Solution

The answer to the question in the problem will be two correct statements. To do this, let's try to solve a few simple problems using a graph and some data. The section of the electrical circuit is a series-connected silver and aluminum wires. A constant electric current of 2 A flows through them. The graph shows how the potential φ changes in this section of the circuit when it is displaced along the wires by a distance x. The specific resistances of silver and aluminum are 0.016 μΩ m and 0.028 μΩ m, respectively.

The wires are connected in series, therefore, the current strength in each section of the circuit will be the same. The electrical resistance of the conductor depends on the material from which the conductor is made, the length of the conductor, the cross-sectional area of ​​\u200b\u200bthe wire

where ρ is the resistivity of the conductor; l- conductor length; S- cross-sectional area. It can be seen from the graph that the length of the silver wire L c = 8 m; aluminum wire length L a \u003d 14 m. Voltage on the section of silver wire U c \u003d Δφ \u003d 6 V - 2 V \u003d 4 V. Voltage in the section of aluminum wire U a \u003d Δφ \u003d 2 V - 1 V \u003d 1 V. According to the condition, it is known that a constant electric current of 2 A flows through the wires, knowing the voltage and current strength, we determine the electrical resistance according to Ohm's law for the circuit section.

It is important to note that the numerical values ​​must be in the SI system for calculations.

Correct statement 2.

Let's check the expressions for power.

P a = I 2 · R a(4);

P a \u003d (2 A) 2 0.5 Ohm \u003d 2 W.

The reference book contains in full the theoretical material on the course of physics, which is necessary for passing the exam. The structure of the book corresponds to the modern codifier of content elements in the subject, on the basis of which the examination tasks are compiled - control and measuring materials (CMM) of the Unified State Examination. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks corresponding to the USE format. This will help the teacher to organize preparation for the unified state exam, and students to independently test their knowledge and readiness for the final exam. At the end of the manual, answers are given to tasks for self-examination, which will help schoolchildren and applicants to objectively assess the level of their knowledge and the degree of preparedness for the certification exam. The manual is addressed to senior students, applicants and teachers.

A small object is located on the main optical axis of a thin converging lens between the focal length and twice the focal length from it. The object is brought closer to the focus of the lens. How does this change the image size and optical power of the lens?

For each quantity, determine the appropriate nature of its change:

1. increases
2. decreases
3. does not change

Write to table selected figures for each physical quantity. Numbers in the answer may be repeated.

Solution

The object is located on the main optical axis of a thin converging lens between the focal and double focal lengths from it. The object begins to be brought closer to the focus of the lens, while the optical power of the lens does not change, since we do not change the lens.

where F is the focal length of the lens; D is the optical power of the lens. To answer the question of how the image size will change, it is necessary to build an image for each position.

Rice. 1

Rice. 2

We built two images for two positions of the subject. It is obvious that the size of the second image has increased.

Answer: 13.

The figure shows a DC circuit. The internal resistance of the current source can be neglected. Establish a correspondence between physical quantities and formulas by which they can be calculated ( - EMF of the current source; R is the resistance of the resistor).

For each position of the first column, select the corresponding position of the second and write in table selected numbers under the corresponding letters.