We knows R∝l Here, R1+R2=12Ω and R1+R2R1×R2=38Ω ⇒R1R2=32Ω We get, R1=8 and R2=4 Again, R1=l1+l212l1 and R2=l1+l212l2 Hence, l2l1=21
A ring is made of a wire having a resistance R_0=12 Ω. Find the points A and B, as shown in the figure, at which a current carrying conductor should…
NEET UG 2012 — Physics Electromagnetism
A ring is made of a wire having a resistance R0=12Ω. Find the points A and B, as shown in the figure, at which a current carrying conductor should be connected so that the resistance R of the sub circuit between these points is equal to 8/3Ω 
Held on 30 Apr 2012 · Verified 9 Jul 2026.
Options
- A
l2l1=85
- B
l2l1=31
- C
l2l1=83
- D
l2l1=21
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