Two wires of resistance R, =(3+1)2 and R= (642)
Q are connected in series, then effective resistance
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Two wires connected in series
R= R1+R2= 25 ohms…….(1)
Again same wire connected in parallel
R=R1R2/R1+R2 =6 ohms………(2)
By equation (1)
R2= 25- R1
Put this in sa (2)
R1(25-R1)/R1+25-R1=6
25R1-R1^2=6×25
R1^2-25R1+150=0
R1= 15ohms or 10 ohms
R2= 10ohms or 15 ohms.
Explanation:
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