Question

Ajutor va rog!!! E urgent!! Dau coroana!!!

Answer 1

$a)P_2=E_2I_2=6*0,5=3 W\\b)I_n=I_1+I_2=>I_1=I_n-I_2=2-0,5=1,5 A\\E_1=I_1r_1+U_n=1,5*1+4,5=1,5+4,5=6 V\\c)t=15 min=900s\\W=U_nI_nt=4,5*2*900=8100 J\\d)E_2=I_2r_2+U_n<=>6=0,5r_2+4,5<=>1,5=0,5r_2=>r_2=3\Omega\\E_p=\frac{\sum\limits^{n}_{i=1} {\frac{E_i}{r_i} } \,}{\sum\limits^n_{i=1} {\frac{1}{r_i} } \, }\\ n=2\\E_p=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2} }{\frac{1}{r_1} + \frac{1}{r_2}}= \frac{\frac{6}{1}+\frac{6}{3} }{\frac{1}{1}+\frac{1}{3} }=\frac{6+2}{\frac{4}{3} }=\frac{8*3}{4}=6 V$

$\frac{1}{r_p}=\sum\limits^n_{i=1} {\frac{1}{r_i} } \, \\ n=2\\\frac{1}{r_p}=\frac{1}{r_1} +\frac{1}{r_2}=\frac{1}{1}+\frac{1}{3}=\frac{4}{3}=>r_p=\frac{3}{4}= 0,75\Omega\\P_{max}=\frac{E_p^2}{4r_p}\\ P_{max}=>R=r_p=>R=0,75\Omega$