# Ohm's law

In the drawing the scheme of the acquaintance is shown you the simplest electric circuit. This closed circuit consists of three elements:

- voltage source – the GB battery;
- the consumer of current – loading of R which can be, for example, filament of the electric lamp or the resistor;
- the conductors connecting the voltage source to loading.

By the way, if to add this chain with the switch, the full scheme of the pocket electric lamp will turn out. R loading having the certain resistance is the site of the chain.

The value of current on this site of the chain depends on tension operating on it and its resistance: the more tension is less than resistance, the big current will go on the site of the chain.

This dependence of current on tension and resistance is expressed by the following formula:

I = U/R, where

- I – the current expressed in amperes, And;
- U – tension in volts, In;
- R – resistance in ohms, Ohm.

This mathematical expression so is read: current on the site of the chain is directly proportional to tension on it and is inversely proportional to its resistance. It is the fundamental law of electrical equipment called the Ohm's law (by the name of G. Oma) for the site of the electric circuit. Using the Ohm's law, it is possible to recognize by two known electrical quantities the unknown the third. Here several examples of practical application of the Ohm's law:

- First example. Tension 25 V. Nado acts on the site of the chain having with a resistance of 5 Ohms to learn value of current on this site of the chain. Decision: I = U/R = 25/5 = 5 A.
- Second example. Voltage of 12 V acts on the site of the chain, creating in it the current equal to 20 mA. What resistance of this site of the chain? First of all current 20 mA needs to be expressed in amperes. It will be 0,02 A. Then R = 12/0,02 = 600 Ohms.
- Third example. Through the site of the chain with a resistance of 10 kOhm current 20 mA flows. What tension operating on this site of the chain? Here, as well as in the previous example, current has to be expressed in amperes (20 mA = 0,02 A), resistance in ohms (10 kOhm = 10000 Ohms). Therefore, U = IR = 0,02×10000 = 200 Century.

On the socle of the glow lamp of the flat pocket lamp vyshtampovano: 0,28 A and 3,5 V.O than tell these data? That the bulb will normally shine at current 0,28 A which is caused of 3,5 V. Using the Ohm's law, it is easy to count that the heated thread of the bulb has R resistance = 3,5/0,28 = 12,5 Ohms.

This resistance to heated bulb thread, resistance to the cooled-down thread is much less. The Ohm's law is fair not only for the site, but also for all electric circuit. In this case in value R the total resistance of all elements of the chain including the internal resistance of the current source is substituted. However at the simplest calculations of chains usually neglect resistance of connecting conductors and internal resistance of the current source.

In this regard it is necessary to give one more example: voltage of electric lighting network is 220 Century. What current will begin to flow in the chain if loading resistance equally in 1000 Ohms? Decision: I = U/R = 220/1000 = 0,22 A. Approximately such current consumes the electric soldering iron.

**All these formulas following from the Ohm's law can use also for calculation of alternating current circuits, but under the condition if in chains there are no inductance coils and condensers.**

It is rather easy to remember the Ohm's law and rated formulas, derivative of it, if to use this graphic scheme, it is the so-called triangle of the Ohm's law.

To use this triangle easy, rather accurately to remember that the horizontal line in it means the division sign (by analogy of fractional line), and the vertical means multiplication sign.

Now it is necessary to consider such question: how does the resistor which is switched on in the chain consistently with loading or parallel to it influence current? It is better to sort it on the example. There is the bulb from round electric, the lamp, expected tension 2,5 B and current 0,075 A. Whether it is possible to feed this bulb from the battery 3336L which initial stress 4,5 B?

It is easy to count that the heated thread of this bulb has resistance a little more than 30 Ohms. If to feed it from the fresh battery 3336L, then through bulb filament, on the Ohm's law, current, almost twice the exceeding that current which it is expected will go. Thread will not sustain such overload, it will be overincandesced and will collapse. But this bulb nevertheless can be fed from the battery 336L if consistently to turn on the additional resistor with a resistance of 25 Ohms in the chain.

In this case the general resistance of the external circuit will be equal about 55 Ohms, that is 30 Ohms – resistance of thread of the bulb of N plus 25 Ohms – resistance of the additional resistor R. In the chain, therefore, the current equal about 0,08 And, that is almost same which bulb filament is expected will begin to flow.

This bulb can be fed from the battery and with more high tension and even from electric lighting network if to pick up the resistor of the corresponding resistance. In this example the additional resistor limits the circuit current to value necessary to us. The more there will be its resistance, the less there will be also the circuit current. In this case the chain included consistently two resistance: resistance of thread of the bulb and resistance of the resistor. And at series connection of resistance current is identical in all points of the chain.

It is possible to turn on the ampermeter in any point, and everywhere it will show one value. This phenomenon can be compared to water flow in the river. The river bed on different sites can be wide or narrow, deep or small. However for the certain period through transverse section of any site of the bed of the river always there passes the identical amount of water.

The additional resistor which is switched on in the chain consistently with loading can be considered as the resistor "extinguishing" the part of tension operating in the chain. Tension which is extinguished by the additional resistor or as speak falls at it, will be that big, than resistance of this resistor is more. Knowing current and resistance of the additional resistor, voltage drop on it is easy to count everything on the same formula U familiar to you = IR, here:

- U – voltage drop, In;
- I – circuit current, A;
- R – resistance of the additional resistor, Ohm.

In relation to the example the resistor R (see fig.) extinguished surplus of tension: U = IR = 0,08×25 = 2 Century. Other tension of the battery equal about 2,5 In, fell to bulb threads. Necessary resistance of the resistor it is possible to find on other formula R familiar to you = U/I where:

- R – required resistance of the additional resistor, Ohm;
- U – tension which needs to be extinguished, In;
- I – circuit current, And.

For the reviewed example resistance of the additional resistor is equal: R = U/I = 2/0,075, 27 Ohms. Changing resistance, it is possible to reduce or increase tension which falls on the additional resistor, thus regulating the circuit current. But the additional resistor R in such chain can be variable, that is the resistor which resistance can be changed (see fig. below).

Regulation of the circuit current by means of the variable resistor. |

In this case by means of the cursor of the resistor it is possible to change smoothly tension brought to load of N so, to smoothly regulate the current proceeding through this loading. The variable resistor which is switched on thus is called the rheostat. By means of rheostats regulate currents in chains of receivers, TVs and amplifiers. At many movie theaters rheostats used for smooth clearing of light in the auditorium. There is also other way of connection of loading to the current source with excess tension – too by means of the variable resistor, but included the potentiometer, that is the voltage divider as shown in the drawing below.

Voltage regulation on R2 loading by means of the variable resistor which is switched on in the electric circuit by the potentiometer. |

Here R1 – the resistor which is switched on by the potentiometer, a R2 – loading which can be the same bulb of incandescence or some other device. On the R1 resistor there is the voltage drop of the current source which partially or can be completely given to R2 loading. When the cursor of the resistor is in extreme lower situation, to loading tension does not move at all (if it is the bulb, it will not burn).

In process of movement of the cursor of the resistor up we will give the increasing tension to R2 loading (if it is the bulb, its thread will be heated). When the cursor of the R1 resistor falls into the extreme upper state, to loading of R2 all tension of the current source will be given (if R2 – the bulb of the pocket lamp, and current source tension big, thread of the bulb fuses). It is possible to find by practical consideration such provision of the cursor of the variable resistor at which to loading tension necessary for it will be given.

The variable resistors which are switched on by potentiometers widely use for regulation of loudness in receivers and amplifiers. The resistor can be directly connected parallel to loading. In that case current on this site of the chain branches and goes two parallel ways: via the additional resistor and the main loading. The greatest current will be in the branch with the smallest resistance.

The sum of currents of both branches will be equal to the current spent for the power supply of the external circuit. Resort to parallel connection in those cases when it is necessary to limit current not in all chain, as at consecutive turning on of the additional resistor but only on some site. Additional resistors connect, for example, parallel to milliammeters that it was possible to measure by them big currents. Such resistors are called shunting or shunts. The word the shunt means the branch.

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