BI 14Bai2. Cho mch in c s nh hnh v, trong ngun sut in ng E= 12 V v in tr trong r = 1,1 , in tr R = 0,1 . 3) in tr x phi c gi tr s l bao nhiu cng sut tiu th mch ngoi l ln nht? 4) in tr x phi c tr s l bao nhiu cng sut tiu th in tr ny l ln nht? Tnh cng sut ln nht .

Gi 3) Cng sut tiu th mch ngoi l P= I2R = M theo bt ng thc Cauchy th :

Du = xy ra khi v ch khi Rtd = r Mt khc: R ni tip vi Rx nn Rtd = R + Rx = r 0.1 + Rx = 1.1

Rx = 1.1 0.1 = 1 Vy Rx = 1 th cng sut tiu th mch ngoi l ln nht 4) Do mch ni tip, cng dng in I nh nhau nn in tr no ln th ta ra cng sut c nhit ln. Vy R cng ln th P cng ln. Do vy cng thc tiu th in tr ny l ln nht th Pmax = Rx I2 =

Rx = Bi 3. Cho mch in c s nh hnh v, trong cc acquy c sut in ng E1 = 12 V, E2 = 6 V v c in tr trong khng ng k. Cc in tr R1 = 4 , R2 = 8.7) Tnh cng sut ca mi acquy v nng lng m mi acquy cung cp trong 5 pht.

Cng dng in ca mch l

I=Cng sut ca mi acquy l P1 = E1I = 12x0.5 = 6W P2 = E2I = 6x0.5 = 3W

Nng lng m mi acquy to ra trong 5 pht l: A1 = P1x t = 6x60x5= 1800 J A2 = P2 x t = 3x60x5 = 900J

BI 15Bi 9 Hai bng n c cc hiu in th nh mc ln lt l U1 = 110 V vU2 = 220 V. Tm t s cc in tr ca chng nu cng sut nh mc ca hai bng c cng sut nh mc ln lt l 25 W v 100 W u lm vic bnh thng hiu in th 110 V. Hi: 10) Cng dng in qua bng n no ln hn? 11) in tr ca bng n no ln hn? 12) C th mc ni tip hai bng n ny vo mng in c hiu in th 220 V c khng? n no s d hng (chy)?

Gai 9) T s in tr ca chng l Khi 2 n sang bnh thng cng mt hiu in th U = 110V th:

R1 = R2 = 10) Cng dng in qua cc n:

I1 =

I2 = Ta c th thy cng dng in n 1 ln hn bng n 2 11) T cu 9 ta c th thy in tr bng n 1 cao hn bng n 2 gp 4 ln 12) Khi gn ni tip vo 2 bng n vo U = 220V. Do mch ni tip th I1 = I2 = It It =

U1 = R1 X I =

U2 = R 2 x I =Ta c th thy U1 = 176 V > U nh mc ca n 1 l 110 V => n 1 s d b h hng Do ta khng c mc ni tip 2 bng n vo hiu in th U = 220V