Question

Dau coroana.
Multumesc anticipat.​

Answer 1

$R_{2,3}=\frac{R_2R_3}{R_2+R_3}=\frac{8*24}{8+24}= \frac{192}{32}= 6\Omega\\R_e=R_1+R_{2,3}=5+6=11\Omega\\R_e=11\Omega$

$I_A=\frac{E}{R_e+r}<=> 1=\frac{12}{11+r}<=>11+r=12=>r=1\Omega$

Legam la bornele bateriei un voltmetru ideal (Rv->+∞) si aplicam a doua  legea lui Kirchhoff pentru ochiul obtinut:

$E=I_Ar+U=>U=E-I_Ar=12-1*1=11 V=>U=11 V$

$R_2,R_3-paralel=>U_2=U_3\\u=I_Ar=1*1=1 V\\U_1=I_AR_1=1*5=5 V\\E=U_1+2U_2+u<=>12=5+1+2U_2<=>6=2U_2=>U_2=3 V\\U_2=R_2I_2=>I_2=\frac{U_2}{R_2}= \frac{3}{8}=0,375A=>I_2=0,375 A$