First slide

Resistors

Question

A ring is made of a wire having a resistance $R_{0} =12 Ω$ . Find the points A and B as shown in fig. at which a current carrying conductor should be connected so that the resistance ‘R’ of the sub circuit between these points is equal to $\frac{8}{3} Ω$

Easy

A

$\frac{l_{1}}{l_{2}} = \frac{5}{8}$
B

$\frac{l_{1}}{l_{2}} = \frac{1}{3}$
C

$\frac{l_{1}}{l_{2}} = \frac{3}{8}$
D

$\frac{l_{1}}{l_{2}} = \frac{1}{2}$

Solution

$L e t s a y R e s i s \tan c e o f l e n g t h l_{1} i s R_{1} a n d R e s i s \tan c e o f l e n g t h l_{2} i s R_{2} I f r i n g i s o p e n, t h e n r e s i s t o r s a r e i n s e r i e s R_{1} + R_{2} = R_{0} = 12 - - - - - (1) I f r i n g i s c l o s e d, t h e n r e s i s t o r s a r e i n p a r a l l e l b e t w e e n A a n d B \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{8}{3} \Rightarrow \frac{R_{1} R_{2}}{12} = \frac{8}{3} \Rightarrow R_{1} R_{2} = 32 A l s o, {(R_{1} - R_{2})}^{2} = {(R_{1} + R_{2})}^{2} - 4 R_{1} R_{2} = {(12)}^{2} - 4 \times 32 = 144 - 128 = 16 \Rightarrow R_{2} - R_{1} = 4 - - - - (2) F r o m (1) a n d (2) w e g e t H e n c e, R_{2} = 8 Ω, R_{1} = 4 Ω ∴ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{4}{8} = \frac{1}{2}$

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