Question

The question in the attachment. Class 9th.

Answer 1

ANSWER:

Option (D) is correct.

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✬ EXPLANATION:

Understanding the Question:

Here's the concept of same potential will be applied.

We are given a resistor circuit and we have to find the equivalent resistance.

Let's start!

Kindly refer to the attachment.

Here I have named the circuit as A,B,C,D,E, and F.

We can see in the given circuit that no resistor is connected between C and D so we will take it as common point as their potential will be same.

The simplified circuit is in the attachment.

R1 and R3 are parellel connection so their equivalent resistance will be:-

$\sf{\to\dfrac{1}{R_{e}}= \dfrac{ 1}{R_{1}}+ \dfrac{1}{R_{2}}}$

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Here the value of each resistance is equal which is R.

$\sf{\to\dfrac{1}{R_{e}}= \dfrac{1 + 1}{R} }$

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$\sf{\to R_{e}=\dfrac{R}{2}}$

Re and R2 are in series connection so equivalent resistance will be:-

$\sf{\to R_{e2}=R_{e} + {R_{2}}}$

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$\sf{\to R_{e2}= \dfrac{R}{2} + {R}}$

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$\sf{\to R_{e2}= \dfrac{R + 2R}{2}}$

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$\sf{\to R_{e2}= \dfrac{3R}{2}}$

$\sf{ R_{e2} \: and \: R_{4} \: are \: in \: parellel \: connection}$

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So Total effective resistance will be:

$\sf {\to\dfrac{1}{R_{p}}=\dfrac{1}R_{e2}+\dfrac{1}{R_{4}}}$

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$\sf {\to\dfrac{1}{R_{p}}=\dfrac{2}{3R}+\dfrac{1}{R}}$

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$\sf {\to\dfrac{1}{R_{p}}=\dfrac{2 + 3}{3R}}$

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$\sf {\to\dfrac{1}{R_{p}}=\dfrac{5}{3R}}$

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$\sf{\to R_{p}=\dfrac{3R}{5}}$

Since equivalent resistance between A and B is 3R/5, option 4 is correct.