Question

Calculează rezistenta echivalenta între bornele A și B pentru rețelele din figură:R1=15 ohm
R2=20 ohm
R3=30 ohm
R4=50 ohm
R5=39ohm
R6=70 ohm
R7=30 ohm
R8=10 ohm
R9=3 ohm
R10=12 ohm
R11=R12=6 ohm
RAB=?
2.R1=R7=4 ohm
R2=R3=R4=R5=R6
=2ohm
RAB=?
Va rog dacă știți ajutați.ma ​

Answer 1

$R_{2,3}=R_2+R_3=20+30=50 \Omega\\R_{2,3,4}=\frac{R_{2,3}*R_4}{R_{2,3}+R_4}=\frac{50*50}{50+50}=\frac{2500}{100}=25\Omega\\R_{1,2,3,4}=R_1+R_{2,3,4}=15+25=40\Omega\\R_{6,7}=70+30=100\Omega\\ R_{5,6,7}=39+100=139\Omega\\R_{11,12}=R_{11}+R_{12}=6+6=12\Omega\\R_{9,10}=\frac{R_9*R_{10}}{R_9+R_{10}}=\frac{3*12}{3+12}=\frac{36}{15} =2,4 \Omega$

$R_{9,10,11,12}=\frac{R_{9,10}*R_{11,12}}{R_{9,10}+R_{11,12}} =\frac{12*2,4}{12+2,4}= \frac{28,8}{14,4}= 2\Omega\\R_{8,9,10,11,12}=R_8+ R_{9,10,11,12}=10+2=12\Omega\\\frac{1}{R_{AB}}=\frac{1}{R_{1,2,3,4}}+\frac{1}{R_{5,6,7}}+\frac{1}{R_{8,9,10,11,12}}=\frac{1}{40}+\frac{1}{139}+\frac{1}{12}= 0,025+0,0071+ 0,083=0,115=>R_{AB}=0,115\Omega$