Question

O sursa avand rezistenta interioara r=0,25Ω disipa pe un rezistor de rezistenta R1=0,01Ω o putere P. Pe ce alt rezistor de rezistenta R2 va disipa sursa aceeasi putere P?

Answer 1

P=I₁²R₁

P=I₂²R₂

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I₂²R₂=I₁²R₁

I₁=E/(R₁+r)

I₂=E/(R₂+r)

E²R₂/(R₂+r)²=E²R₁/(R₁+r)²|:E²=>R₂(R₁+r)²=R₁(R₂+r)²<=>R₂R₁²+R₂r²+2R₁R₂r=R₁R₂²+R₁r²+2R₁R₂r<=>R₂R₁²+R₂r²=R₁R₂²+R₁r²<=>R₁R₂²-R₂(R₁²+r²)+R₁r²=0

Δ=[-(R₁²+r²)]²-4R1²r²=(R₁²+r²)²-4R1²r²=(0,0001+0,0625)²-4*0,0001*0,0625=0,003918-0,000025=0,003893

R₁₂={(R₁²+r²)±√0,003893]/2R₁=(0,0626±0,0623)/0,02=3,12±3,11

R₁=3,12-3,11=0,01 Ω

R₂=3,12+3,11=6,23  Ω