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Standard XII

Physics

Series and Parallel Combination of Resistors

A ring is mad...

Question

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 be connected so that the resistance $R$ of the sub-circuit between these points is equal to $\frac{8}{3} Ω$ .

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

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

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

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

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Solution

The correct option is D $\frac{l_{1}}{l_{2}} = \frac{1}{2}$
Ring is divided into $l_{1}$ and $l_{2}$ .
Let resistance is $R_{1}$ for length $l_{1}$ and $R_{2}$ for length $l_{2}$ .
Now, as per the problem
$\frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{8}{3}$ and
$R_{1} + R_{2} = 12 \Rightarrow R_{1} R_{2} = 32$
Now, we know that
$R_{1} - R_{2} = \sqrt{(R_{1} + R_{2})^{2} - 4 R_{1} R_{2}}$
$= \sqrt{12^{2} - 4 \times 32} = 4 Ω$

So, $R_{1} = 4 Ω a n d R_{2} = 8 Ω$
Hence, $\frac{l_{1}}{l_{2}} = \frac{R_{1}}{R_{2}} = \frac{4}{8} = \frac{1}{2}$

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A ring is made of a wire having a resistance R0=12 Ω. 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 83 Ω.

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 be connected so that the resistance $R$ of the sub-circuit between these points is equal to $\frac{8}{3} Ω$ .