Državni izpitni center ELEKTROTEHNIKA. Izpitna pola 1. Četrtek, 1. junij 2017 / 90 minut
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1 Š i f r a k a n d i d a t a : Državni izpitni center *M * SPOMLADANSKI IZPITNI ROK ELEKTROTEHNIKA Izpitna pola 1 Četrtek, 1. junij 017 / 90 minut Dovoljeno gradivo in pripomočki: Kandidat prinese nalivno pero ali kemični svinčnik, svinčnik, radirko, šestilo, trikotnika in računalo. Kandidat dobi dva konceptna lista in ocenjevalni obrazec. Priloga s konstantami in enačbami ter magnetilnimi krivuljami je na perforiranem listu, ki ga kandidat pazljivo iztrga. SPLOŠNA MATURA NAVODILA KANDIDATU Pazljivo preberite ta navodila. Ne odpirajte izpitne pole in ne začenjajte reševati nalog, dokler vam nadzorni učitelj tega ne dovoli. Prilepite kodo oziroma vpišite svojo šifro (v okvirček desno zgoraj na tej strani in na ocenjevalni obrazec). Svojo šifro vpišite tudi na konceptna lista. Izpitna pola vsebuje 8 nalog s kratkimi odgovori in 3 strukturirane naloge. Število točk, ki jih lahko dosežete, je 40. Za posamezno nalogo je število točk navedeno v izpitni poli. Pri reševanju si lahko pomagate z zbirko konstant in enačb v prilogi. Rešitve, ki jih pišite z nalivnim peresom ali s kemičnim svinčnikom, vpisujte v izpitno polo v za to predvideni prostor, slike in diagrame pa rišite prostoročno s svinčnikom. Pišite čitljivo. Če se zmotite, napisano prečrtajte in rešitev zapišite na novo. Nečitljivi zapisi in nejasni popravki bodo ocenjeni z 0 točkami. Osnutki rešitev, ki jih lahko naredite na konceptna lista, se pri ocenjevanju ne upoštevajo. Zaupajte vase in v svoje zmožnosti. Želimo vam veliko uspeha. Ta pola ima 16 strani, od tega prazni. Državni izpitni center Vse pravice pridržane.
2 /16 *M *
3 *M * 3/16 Konstante in enačbe Elektrina in električni tok 19 e 0 = 1,60 10 C Q = ( ± ) ne0 DQ i = D t I = JA m = cit Električno polje 1 ε As 0 = 8, Vm QQ 1 F = 4 πε d F = QE Q E = 4 πε r q E = πε r E = σ ε D = εe = εε 0 re U = Ed UAB = VA VB D e = Q = DA Q C = C = ε A U d W = CU w = ED Enosmerna vezja ( ± ) I = 0 k m k ( ± ) U = 0 m R = U = 1 I G P = UI W = Pt ρl R = = l A γ A Rϑ = 1+ α ( ϑ 0 C) R0 Pizh h = P vh Magnetno polje 7 m Vs 0 = 4π 10 Am miil 1 F = p d F = BIl F = B A µ o µ I B = π r µ Ir B = π r0 µ NI B = l F = BA M = IAB sinα Θ= Hl B = µ H = µµ 0 rh R l m = m A Inducirano električno polje Y = N F u i = Y t ui = vbl Um = ωnf m L = Y µ N A L = i l W = Li w = BH F = B A µ 0 Trifazni sistemi Y1U + YU + Y3U V 0 = Y + Y + Y Izmenična električna vezja ω = π f Tf = 1 ( ω α ) ( ω α ) u = U sin t + u i = I sin t + i j = a - a jα u e = cosα + jsinα U Z = = 1 I Y Z = R+ jx Y = G+ jb ZR Z L Z C = R = jωl = 1 jωc S= P+ jq = UI Q tanδ = 1 0 LC 1 ω = ω0l Q = = 1 R ω CR Prehodni pojavi u = Ri u = L di d t i = C du d t ( u U 1 e t / t = ) u = Ue t /t τ = RC ( i I 1 e t / t = ) i = Ie t /t τ = L R i 0 P perforiran list
4 4/16 *M *
5 *M * 5/16 1. Dnevna poraba električne energije nekega gospodinjstva je 1 kwh. Koliko MJ ustreza tej energiji?. Dielektričnost in permeabilnost vakuuma sta osnovni elektrotehniški konstanti. Napišite enačbo, ki povezuje ti dve konstanti s svetlobno hitrostjo.
6 6/16 *M * 3. Curek elektronov, ki udarja na zaslon osciloskopa, prinese vsako μs 3,5 milijona elektronov. Kolikšen je električni tok v takšnem curku? 4. Na enosmerni vir U = 4 V je priključeno breme. Upornost vodnikov, ki povezujejo vir in breme, je R v = 0,3 Ω. Notranja upornost vira je R n = 0,7 Ω, upornost bremena pa R b = 5 Ω. R v R n I + U R b Izračunajte tok skozi breme.
7 *M * 7/16 5. Izkoristek električnega bremena je: A razmerje med opravljenim koristnim delom in sprejeto električno energijo, B razmerje med sprejeto električno in oddano koristno močjo, C razmerje med toplotnim in mehanskim delom. Obkrožite črko pred pravilnim odgovorom. 6. Pri frekvenci f 1 = 10 khz je reaktanca tuljave X 1 = 1000 Ω. Izračunajte reaktanco iste tuljave pri frekvenci f = 0 khz. L
8 8/16 *M * 7. Harmonični napetosti pripada kazalec efektivne vrednosti U = ( 10 j10) V. Izračunajte amplitudo U m. 8. Zaporedno vezavo upora R = 0 Ω in tuljave L = 0,1 H priključimo na vir enosmerne napetosti U = 10 V. Izračunajte tok i tuljave v trenutku t = 5 ms po priključitvi vezja na napetost.
9 *M * 9/16 Prazna stran OBRNITE LIST.
10 10/16 *M * 9. Dana je UI-karakteristika žarnice z žarilno nitko do nazivne napetosti U n = 4 V UI karakteristika žarnice I [ ] ma U [ V] 9.1. Kolikšen je tok I skozi žarnico pri napetosti 1 U 1 = 1 V? 9.. Kolikšna je moč žarnice P pri napetosti U = 15 V?
11 *M * 11/ Koliko električne energije W prejme žarnica pri nazivni napetosti v času t = 5 h? 9.4. Za koliko odstotkov se zmanjša moč žarnice, če se napetost iz nazivne zmanjša za polovico?
12 1/16 *M * 10. Kompleksno breme z admitanco Y = ( 4 j3 ) S je priključeno na harmonični tokovni vir z amplitudo I m = 10 A in krožno fekvenco ω = 400 Hz Izračunajte impedanco bremena Izračunajte amplitudo napetosti na bremenu.
13 *M * 13/ Izračunajte kompleksno moč bremena Izračunajte kapacitivnost kondenzatorja, ki bo v celoti kompenziral jalovo moč bremena.
14 14/16 *M * 11. Podatki vezja so: U = 4 V, R = 10 Ω in L = 100 mh. V času t = 0 s sklenemo stikalo. + - R U L R Izračunajte tok skozi tuljavo pred sklenitvijo stikala Izračunajte magnetno energijo v tuljavi po končanem prehodnem pojavu.
15 *M * 15/ Narišite časovni diagram toka skozi tuljavo po sklenitvi stikala Kolikšna je napetost med priključkoma tuljave tik po sklenitvi stikala?
16 16/16 *M * Prazna stran
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