Question

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Answer 1

☞ $\displaystyle\large\underline{\sf\red{Given}}$

✭ There are two resistors of 1Ω each connected in parallel

✭ These are then connected to a resistor of 1Ω & 5Ω each(in parallel)

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ Net resistance of the combination?

$\displaystyle\large\underline{\sf\gray{Solution}}$

Series connection of resistors is given by,

$\displaystyle\sf \underline{\boxed{\sf R_{net} = R_1 + R_2+R_3...R_n}}$

Also Parallel Connection of resistance is given by,

$\displaystyle\sf \underline{\boxed{\sf\dfrac{1}{R_{net}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}...\dfrac{1}{R_n}}}$

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

There is a combination of 2 resistors of 1Ω & 2Ω resistors connected in parallel so their equivalent resistance will be,

$\displaystyle\sf\twoheadrightarrow\dfrac{1}{R_{net}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}...\dfrac{1}{R_n}$

• R_1 = 1Ω
• R_2 = 2Ω

Substituting the values,

$\displaystyle\sf\twoheadrightarrow\dfrac{1}{R_{eq}} = \dfrac{1}{1} + \dfrac{1}{2}$

$\displaystyle\sf\twoheadrightarrow\dfrac{1}{R_{eq}} = \dfrac{1}{1} \times \dfrac{2}{2} +\dfrac{1}{2}$

$\displaystyle\sf\twoheadrightarrow\dfrac{1}{R_{eq}} = \dfrac{2}{2} \times \dfrac{1}{2}$

$\displaystyle\sf\twoheadrightarrow\dfrac{1}{R_{eq}} = \dfrac{2+1}{2}$

$\displaystyle\sf\twoheadrightarrow\dfrac{1}{R_{eq}} = \dfrac{3}{2}$

$\displaystyle\sf\twoheadrightarrow \orange{R_{eq} = \dfrac{2}{3} \Omega}$

So now these three resistors are in series and so their equivalent resistance gives us our net resistance, which is,

$\displaystyle\sf\dashrightarrow R_{net} = R_1 + R_2+R_3...R_n$

• R_1 = 1Ω
• R_2 = 2Ω
• R_3 = ⅔Ω

$\displaystyle\sf\dashrightarrow R_{net} = 1+1+\dfrac{2}{3}$

$\displaystyle\sf\dashrightarrow R_{net} = 2+\dfrac{2}{3}$

$\displaystyle\sf\dashrightarrow R_{net} = \dfrac{6+2}{3}$

$\displaystyle\sf\dashrightarrow R_{net} = \dfrac{8}{3}$

$\displaystyle\sf\dashrightarrow\pink{ R_{net} = 2.7 \ \Omega}$

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