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sergeysargsyanotl7bg
@sergeysargsyanotl7bg
August 2022
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Найти общий сопротивление и мощность 4 лампы . Если
R1=2ом
R2=1ом
R3=5ом
R4=1 ом
I = 2 А
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fatt52
Verified answer
R12=R1+R2=2+1=3 Ом
R34=R3+R4=5+1=6 Ом
Ro=R12*R34/(R12+R34)=3*6/(3+6)=2 Ом
Uo=I*Ro=2*2=4 B
I4=Uo/R34=4/6=0.67 A
P4=I4²*R4=0.67²*1=0.44 Вт
===========================
2 votes
Thanks 1
ilyusha2145
Verified answer
R1=R1+R2=2 Om+1 Om=3 Om
R2=R3+R4=5 Om+1 Om=6 Om
R=R1*R2/R1+R2=18 Om/9 Om=2 Om.
U=IR=4V
I=U/R=4v/6 Om=0,67A
P=UI=I^2*R=0,4489*1 Om=0,4489Вт.
1 votes
Thanks 0
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Answers & Comments
Verified answer
R12=R1+R2=2+1=3 ОмR34=R3+R4=5+1=6 Ом
Ro=R12*R34/(R12+R34)=3*6/(3+6)=2 Ом
Uo=I*Ro=2*2=4 B
I4=Uo/R34=4/6=0.67 A
P4=I4²*R4=0.67²*1=0.44 Вт
===========================
Verified answer
R1=R1+R2=2 Om+1 Om=3 OmR2=R3+R4=5 Om+1 Om=6 Om
R=R1*R2/R1+R2=18 Om/9 Om=2 Om.
U=IR=4V
I=U/R=4v/6 Om=0,67A
P=UI=I^2*R=0,4489*1 Om=0,4489Вт.