Question

Un generator debitează aceeași putere P=100W.Fie că la bornele sale se leagă R1=4omi sau R2=9omi.Se cere:
a)Tensiunea electromotoare a generatorului(E=?)
b)Rezistența internă(r=?)
c)Randamentele circuitului(n1,n2=?)
VĂ ROGGGGGGGG!!!
E URGENT!!!!

Answer 1

$I_1=\frac{E}{R_1+r}\\ I_2=\frac{E}{R_2+r}\\ P_1=R_1I_1^2=\frac{R_1E^2}{(R_1+r)^2}=>E=\frac{P_1(R_1+r)^2}{R_1}\\ P_2=R_2I_2^2=\frac{R_2E^2}{(R_2+r)^2}=\frac{R_2\frac{P_1(R_1+r)^2}{R_1}}{(R_2+r)^2} \\P_1=P_2=>(R_2+r)^2R_1=R_2(R_1+r)^2<=>R_1R_2^2+2R_1R_2r+r^2R_1=R_1^2R_2+2R_1R_2r+r^2R_2<=>R_1R_2(R_2-R_1)=$

$=r^2(R_2-R_1)|:(R_2-R_1)=>r=\sqrt{R_1R_2}=\sqrt{4*9}=2*3=6\Omega=>r=6\Omega\\E=\frac{P(R_1+r)^2}{R_1}=\frac{100(4+6)^2}{4}=25*100=2500 V=>E=2500V\\ \eta_1=\frac{R_1}{R_1+r}=\frac{4}{4+6} =\frac{4}{10}=0,4=>\eta_1=40\%\\ \eta_2=\frac{R_2}{R_2+r}=\frac{9}{9+6}=\frac{9}{15}=0,6=>\eta_2=60\%$