Question

please solve this question​

Answer 1

Step-by-step explanation:

In this question, we are asked to find the equivalent resistance between two points in a circuit of three resistors. In order to solve this question, we must have a basic knowledge of series and parallel connection of resistors. When resistors are connected from end to end, they are said to be in a series connection but when ending point of two or more resistors are connected to a single point, they are said to be in parallel connection.

In the given circuit, we can see that 4Ω and 6Ω resistors are in parallel connection as per the definition we discussed above.

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Assume that :-

• $\sf R_1$ = 4Ω

• $\sf R_2$ = 6Ω

• $\sf R_3$ = 2Ω

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Applying formula to find equivalent resistance in parallel connection :

${\sf : \implies \dfrac{1}{R_{(P)}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}$

${\sf : \implies \dfrac{1}{R_{(P)}} = \dfrac{1}{4\Omega} + \dfrac{1}{6\Omega}}$

${\sf : \implies \dfrac{1}{R_{(P)}} = \dfrac{3 + 2}{12\Omega}}$

${\sf : \implies \dfrac{1}{R_{(P)}} = \dfrac{5}{12\Omega}}$

Cross multiply

$\boxed{ \purple{\sf : \implies R_{(P)} = \dfrac{12\Omega}{5} }}$

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Now we can see that this combination of resistors $\sf R_{(P)}$ is in series connection with 2 Ω resistor.

Applying formula for resistors in series connection:

${\sf : \implies R_{(e)} =R_{(3)} + R_{(P)} }$

${\sf : \implies R_{(e)} = 2 \Omega+\dfrac{12\Omega}{5} }$

${\sf : \implies R_{(e)} = \dfrac{10\Omega + 12\Omega}{5} }$

${\sf : \implies R_{(e)} = \dfrac{22\Omega}{5} }$

$\boxed{ \blue{\sf : \implies R_{(e)} =4.4\Omega } }$

So the equivalent resistance of whole circuit is 4.4Ω.

[Omega will change into Ω when viewed from website].