DC circuits with a single source

Transkript

1 Název projektu: utomatizace výrobních procesů ve strojírenství a řemeslech egistrační číslo: Z..07/..0/0.008 Příjemce: SPŠ strojnická a SOŠ profesora Švejcara Plzeň, Klatovská 09 Tento projekt je spolufinancován Evropskou unií a státním rozpočtem České republiky Produkt: Zavádění cizojazyčné terminologie do výuky odborných předmětů a do laboratorních cvičení D circuits with a single source Návod v anglickém jazyce Číslo tématu: b Monitorovací indikátor:

2 NSTTONS FO TOP: b reated in school year: 0/0 Branch: 6-4-M/0 Electrical Engineering - Mechatronics Subject: Electrical Engineering Year:. Prepared by: ng. Jitka oubalová; translated by: Mgr. Marie Mádlová D circuits with a single source This topic deals with electrical D circuits with a single source of D voltage and resistive load. The load is made up by several resistors that are connected in series, parallelly, or combined through different topographies. When solving this task, we have to reduce the load to one single resistor. Then applying Ohm s Law we can determine the value of current taken from the source. Then we gradually convert the simple circuit back up to the initial resistors connections. Now we are able to derive the currents in each of the circuit branches and the voltage of each element of the circuit. For better understanding here are some examples and solutions. Example : What is the value of the current taken from the source and the voltage on resistors and when we have the circuit as in the fig. fig. First we replace the load by only one resistor, it must get the same effect as in the initial circuit.

3 esistors are connected in series, so the resulting resistance is, = +. See the new auxiliary circuit, urrent taken from the source is :, Now we convert the circuit back to the initial circuit and calculate the voltage of each resistor seperately. Example : What is the value of the current taken from the source,the voltage of resistors, a and currents flowing in the branches when we have the circuit as in fig. B Fig. B

4 esistors are connected parallelly, therefore if we replace them by, its resistance is, so We get the following simple circuit,, B The result resistor,, = +,. So the auxiliar circuit is :,, nd we can calculate the current : Now we convert the circuit back to the initial circuit and gradually calculate voltages of each resistor and currents in each circuit branches :,, B B 4

5 B B Note: We can solve such and similar examples by more than one possible way of solutions to get the values of voltage and currents. Here is another procedure how to solve it and get the same and correct results: The first step: simplifying the circuit and calculating the total current and voltage B,, Next step is different and also possible : B The result is the same for both procedures., B Example : What are the current and voltage values of all elements of the circuit in the fig.? The values of the circuit elements are: =48V, =Ω, =5Ω, =8Ω, 4 =Ω, 5 =6Ω. + _ B fig. 5

6 We solve this circuit by the gradual simplifying of the connected resistors. First we derive the resistance of parallely connected resistors 4 and 5, then we draw the simplified diagram. Following formula is applied: 45 The simplified circuit is : _ B 45 esistors and 45 are connected in series, resulting resistence is Now we get the simplified diagram + _ 45 We substitude resistors and 45 that are parallely connected by resistor 45,and we calculate its value using following Here is the auxiliary circuit 6 45

7 + _ 45 esistors and 45 are connected in series, the total result resistance of the auxiliary circuit is The total current taken from the source is Voltage of the resistor is following: 6 V Voltage between nodes and is following: 66 6V 45 alculate the currents and : 6, 4 5 6, alculate voltage values of the resistors and using the values of currents a : 5,4 6V 8,6 8, 8V Now calculate the voltage between nodes B and : B 4 5 B,6 7, V 45 7

8 From the value of voltage B we derive the currents value 4 and 5 in resistors 4 a 5 : B 7, 4, 4 4 B 7, 5, 6 5 This is the final result. Note.: We can check the results by approving the currents of the nodes and voltage of the loops applying Kirchhoff s Laws. You have to apply.kirchhoff s Law for the node : 6,4,6 0 For the node B apply: 0 4 5,6,4, 0 For the node apply: 0 4 5,4,4, 6 0 You have to apply. Kirchhof s Law for the loop x: 0, For the loop y apply the following: 4 0 0, 8,8 7, 6 0. esults are approved by Kirchhoff s laws. x y 8

9 Řešení obvodů stejnosměrného proudu s jedním zdrojem - D circuits with a single source - slovníček odborných termínů Vocabulary Slovníček apply approve auxiliary calculate circuit connect, connection platit, používat ověřit pomocný vypočítat obvod zapojit, zapojení current D proud, stejnosměrný derive D (voltage, circuit) flow, the current flows in series initial Kirchhoff s Law load loop node note Ohm s Law parallel resistence, resistor resistive simplify source substitude value voltage odvodit, vypočítat stejnosměrný téci za sebou, seriově počáteční, původní Kirchhoffův zákon zátěž smyčka uzel poznámka Ohmův zákon paralelní, vedle sebe odpor odporový zjednodušit zdroj nahradit hodnota napětí Zdroj: OBLOVÁ, J., Elektrotechnika [online]. [cit ]. Dostupné z WWW: < 9