Ohm's law

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Ohm’s law
Y.A.SMORODINSKY

Simple at first glance

Typically Ohm’s law is formulated this way. Let’s take a conductor with current (Fig. 1). Let a potential difference $V_A - V_B$ be applied to its ends, and $V_A > V_B$. then the strength of the current flowing through the conductor will be determined by the formula

$$I = \frac{V_A - V_B}{R},$$

where $R$ is the conductor resistance. In addition, they add that if a conductor with a constant cross-section $S$ is uniform along the entire length $l$, then $R = \frac{1}{\sigma} \frac{l}{S}$, where $\sigma$ is called specific conductivity, and $\frac{1}{\sigma} = \rho$ is resistivity.

The graph of the dependence of the current on the applied potential difference is expressed as a straight line (Fig. 2).

Ohm’s law looks very simple and very powerful. After all, it is clear that the greater the potential difference,

$$V_A \quad V_B \quad I$$

Fig. 1.

tem more and current. But why is there a simple proportionality between these two quantities, why does the quantity $R$ itself not change with increasing current? Experience has shown that Ohm’s law is true for both very small and very large currents in metals. Such a large potential difference was applied to the wiring that over a length of 1 cm the potential changes by a million volts. And yet the current could be determined using Ohm’s law. True, in fact, in such experiments very small deviations from this law were observed, but they could be explained by the fact that the conductor, despite all precautions, heated up, and its resistance changed due to changes in temperature. This seemingly simple law raises interesting questions.

In which direction does the current flow?

Why does current flow in the direction of decreasing potential, and not in the direction of increasing it? Let’s take a battery and connect it to some resistance, for example, turn on a light bulb. According to Ohm’s law, the circuit will consume current from the positive terminal of the battery through the light bulb to the imprinted terminal, reducing the potential difference between the terminals, which of course will be restored by the battery, but this consumes energy all the time (this will be discussed later).

What would happen if the current went in the opposite direction? Terminal charges

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is expressed temporally by an exponential function *).

Knowing the properties of the exponential function, we can write that the charge changes as follows:

$$Q(t) = Qe^{-\frac{2t}{CR}}$$

Thus, the charge will decrease according to an exponential law. Technically speaking, the charges will disappear only after an infinitely long time, but in practice they will become invisible very quickly.

Typically, the time for establishing equilibrium is called the time during which the charges of the balls (and therefore the current in the circuit) decreases by $e$ times. For our example, this time is $\frac{CR}{2}$.

History of discovery

We still consider Ohm’s law to be one of the most important laws of physics. When Georg Simon Ohm (1787-1854) discovered this law, they simply did not believe him. Ohm’s main work was published in 1927 in Berlin and was called “Galvanic goal, mathematical developed by the doctor.”

*) The exhibitor is described in «Kvant» No. 1 for this year. Only there α is chosen equal to one.

**) Let’s check the last formula. Let’s consider two moments of time $t$ and $t+\Delta t$.

$$Q(t+\Delta t) = Qe^{-\frac{2}{CR}(t+\Delta t)}$$

$$Q(t) = Qe^{-\frac{2}{CR}t}$$

The difference between these two functions can be calculated (for small $\Delta t$). Note that we are talking about changing the function $e^x$, and $x$ changes from $\left(-\frac{2}{CR}t\right)$ to $\left[-\frac{2}{CR}(t+\Delta t)\right]$, that is, by values ​​$\Delta x = -\frac{2}{CR}\Delta t$.

Thus, $\Delta Q = -\frac{2}{CR}\Delta t \cdot Q$, since the rate of change of an exponential function is proportional to itself.

G.S. Ohm." But his first research began earlier; their results were published in 1825. To understand how difficult it was to achieve recognition of these works, you need to know what tools an experimental physicist had at his disposal 150 years ago.

In 1820, Ampere discovered that an electric current deflects the magnetic needle of a compass. This allowed us to basically change the “current strength” and understand that there is a difference between current strength and voltage. By this time, two current sources were known: the electric battery “voltaic” (invented in 1799 by A. Volta) and the thermodynamic element invented by Seebeck in 1821. However, he did not foresee that it was possible to establish mathematical laws of battery operation. Physicists were surprised at the change in the action of the battery with different connecting elements, but they could not see any clear patterns here. In addition, galvanic batteries at that time were flat, the direction at their ends changed during the experiment (the batteries “sat down”), which greatly confused the results of the experiments. Ohm abandoned these batteries and conducted his first experiments using a thermodynamic element as a current source, which gave a significantly lower current, but the voltage was stable. The potential difference of such an element is about 0.0045 volts. To change the current obtained from thermal action, Ohm came up with a new method. The current-carrying conductor was positioned along the direction of a freely suspended magnetic needle (i.e., along the meridian), and the angle at which the thread had to be twisted changed so that the needle did not deviate from this direction. The angle of twist of the thread characterized the current strength.

The first formula that Ohm received was of the form

$$X = \frac{a}{b+x},$$

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acceleration $\frac{e}{m} \vec{E}$ multiplied by τ:
$$\vec{u} = \frac{e}{m} \vec{E} \tau.$$

If the number of electrons (“free electrons,” that is, those that are not “attached” to atoms) per unit volume is denoted by $N$, then an electric current will flow through $1 \, cm^2$ equal to the charge of the electron multiplied by the number of electrons passing through $1 \, cm^2$ in 1 second

$$\vec{j} = \frac{1}{2} \frac{e^2}{m} N \tau \vec{E}.$$

In this formula, together with the speed $u$ we wrote half of it. It is simply the arithmetic average of the initial and final velocities between two collisions.

From here we can write an expression for conductivity (as a proportionality coefficient between $\vec{j}$ and $\vec{E}$):

$$\sigma = \frac{1}{2} \frac{e^2 N \tau}{m}.$$

Joule heat and current source work

Thus, behind the simple formula of Ohm’s law there are many complex phenomena hidden. When current flows, electrons constantly receive energy from the electric field and always lose it when interacting with atoms. As a result, as is well known, heat is released

*) Sometimes the same formula is written slightly differently, replacing τ by the formula

$$\tau = \frac{\text{свободный пробег}}{\text{средняя тепл. скорость}} = \frac{\lambda}{v_{\text{cp}}}.$$

The mean free path is the average path taken by an electron between two collisions. Then

$$\sigma = \frac{1}{2} \frac{e^2 N \lambda}{m v_{\text{cp}}}.$$

This formula is called the Drude formula.

(for example, an electric stove heats up, the wire in which has a high resistance).

The amount of heat that is released in a section of a conductor with resistance is determined by the Joule-Lenz formula

$$Q = \frac{1}{J} I^2 R,$$

where $J$ is the mechanical equivalent of heat:

$$J = 4,184 \frac{\partial \chi}{\kappa \cdot \kappa a l}.$$

What resources generate heat? Answer: just as the current source does this work, it maintains the potential at its terminals.

Returning to Ohm’s formula for a closed circuit, we can say that the energy expended in the form of heat in the entire circuit, including the interior of the current source itself, is equal to $E I$, that is, the product of E. d. c. on current strength

$$I^2 (R + r) = E I.$$

The work done by a current source occurs either due to chemical energy (in batteries) or due to mechanical energy (in generators). The last formula expresses the law of conservation of energy for a circuit consisting of a current source closed by some resistance.

The formula of Ohm’s law allows you to calculate any direct current circuit, and in a slightly modified form it can be generalized to alternating currents.

Exercises

1. Find by what law the potentials of two balls similar to those drawn on figure 3 will change if the charge of one is $Q=0,01$ coulomb and the other is 0.02 coulomb. Set the product $CR$ equal to $10^{-12}$ sec.

2. After how long will the difference between the charges of the balls decrease by 20 times?