when two resistors of resistance R1 and R2 they are connected in parallel the net resistance is 3 ohm when they are connected in series the value 16 ohm calculate R1 and R2
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It can be solved by substitution method:
In series; R1 +R2 =16 《R2=16-R1》
1/R1 +1/R2=1/3
(R2+R1)/R1R2 =1/3
16/R1R2 =1/3
R1R2= 48
substituting R2=16-R1
16/R1 (16-R1)= 1/3
16/16R1-R1^2 = 1/3
R1^2-16R1+48 =0
•°•by splitting method
R1^2 -12R1 -4R1 +48 =0
R1 (R1-12) -4 (R1-12) =0
(R1-4) (R1-12) =0
•°• R1=12 or R1=4
Hence, two resistors can be 4 ohm or 12 ohm
In series; R1 +R2 =16 《R2=16-R1》
1/R1 +1/R2=1/3
(R2+R1)/R1R2 =1/3
16/R1R2 =1/3
R1R2= 48
substituting R2=16-R1
16/R1 (16-R1)= 1/3
16/16R1-R1^2 = 1/3
R1^2-16R1+48 =0
•°•by splitting method
R1^2 -12R1 -4R1 +48 =0
R1 (R1-12) -4 (R1-12) =0
(R1-4) (R1-12) =0
•°• R1=12 or R1=4
Hence, two resistors can be 4 ohm or 12 ohm
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