Question

What is the equivalent resistance between points P and Q in the following network?
With full explanation. the explained answer will receive the tag of brainliaist.
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Answer 1

$\bigstar\:{\underline{\sf Step \: 1 \::}}$

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$\sf Refer\: to\: the\: attachment$

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$\bigstar\:{\underline{\sf Step \: 2 \::}}$

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$\sf \implies R_1 \: and R_2\: are\: connected\: in \: series$

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$\star\;{\boxed{\sf{\pink{R' = R_1 + R_2 }}}}$

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$:\implies\sf R' = 2Ω + 2Ω$

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$:\implies\sf R' = 4Ω$

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$:\implies{\underline{\boxed{\frak{{R'= 4Ω }}}}}\;\bigstar$

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$\bigstar\:{\underline{\sf Step \: 3 \::}}$

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$\sf \implies R' \: and R_5\: are \: connected\: in \: parallel$

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$\star\;{\boxed{\sf{\pink{\dfrac{1}{R''} = \dfrac{1}{R'} + \dfrac{1}{R_5} }}}}$

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$:\implies\sf \dfrac{1}{R''} = \dfrac{1}{4} + \dfrac{1}{4}$

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$:\implies\sf \dfrac{1}{R''} = \dfrac{2}{4}$

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$:\implies\sf \dfrac{1}{R''} = \dfrac{1}{2}$

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$:\implies{\underline{\boxed{\frak{{R''= 2Ω }}}}}\;\bigstar$

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$\bigstar\:{\underline{\sf Step \: 4 \::}}$

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$\sf \implies R''\: and R_6\: are\: connected\: in \: series$

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$\star\;{\boxed{\sf{\pink{R''' = R'' + R_6 }}}}$

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$:\implies\sf R''' = 2Ω + 2Ω$

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$:\implies\sf R''' = 4Ω$

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$:\implies{\underline{\boxed{\frak{{R'''= 4Ω }}}}}\;\bigstar$

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$\bigstar\:{\underline{\sf Step \: 5 \::}}$

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$\sf \implies R''' \: and \: R_7\: are \: connected\: \: in\: parallel$

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$\star\;{\boxed{\sf{\pink{\dfrac{1}{R''''} = \dfrac{1}{R'''} + \dfrac{1}{R_7} }}}}$

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$:\implies\sf \dfrac{1}{R''''} = \dfrac{1}{4} + \dfrac{1}{4}$

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$:\implies\sf \dfrac{1}{R''''} = \dfrac{2}{4}$

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$:\implies\sf \dfrac{1}{R''''} = \dfrac{1}{2}$

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$:\implies{\underline{\boxed{\frak{{R''''= 2Ω }}}}}\;\bigstar$

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$\bigstar\:{\underline{\sf Step \: 6 \::}}$

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$\sf \implies R''''\: and R_4\: are\: connected\: in \: series$

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$\star\;{\boxed{\sf{\pink{R''''' = R'''' + R_4 }}}}$

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$:\implies\sf R''''' = 2Ω + 2Ω$

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$:\implies\sf R''''' = 4Ω$

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$:\implies{\underline{\boxed{\frak{{R'''''= 4Ω }}}}}\;\bigstar$

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$\bigstar\:{\underline{\sf Step \: 7 \::}}$

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$\sf \implies R'''''\: and\: R_3 \: are \: connected\: in\: parallel$

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$\star\;{\boxed{\sf{\pink{\dfrac{1}{R} = \dfrac{1}{R'''''} + \dfrac{1}{R_3} }}}}$

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$:\implies\sf \dfrac{1}{R} = \dfrac{1}{4} + \dfrac{1}{4}$

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$:\implies\sf \dfrac{1}{R} = \dfrac{2}{4}$

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$:\implies\sf \dfrac{1}{R} = \dfrac{1}{2}$

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$:\implies{\underline{\boxed{\frak{{R= 2Ω }}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{ Total\: resistance \: across \: P\: and\: Q \: \bf{ 2Ω}}}}$

Answer 2

$\\ {\bf{To\: Find\::}}$

• The equivalent resistance between points P & Q in the given circuit .

$\\ {\bf{Solution\::}}$

➣ See the attachment diagram for numbering of resistors.

Case - 1 :

✯ 1st & 2nd resistor are connected in series.

Thus,

➠ $\tt{R_{eq_1}\:=\:2\Omega\:+\:2\Omega\:}$

➠ $\bf{R_{eq_1}\:=\:4\Omega\:}$

Case - 2 :

✯ The equivalent resistance of 1st & 2nd resistor is connected parallel to 3rd resistor.

Thus,

➠ $\tt{\dfrac{1}{R_{eq_2}}\:=\:\dfrac{1}{4\Omega}\:+\:\dfrac{1}{4\Omega}\:}$

➠ $\tt{\dfrac{1}{R_{eq_2}}\:=\:\dfrac{2}{4}\Omega\:}$

➠ $\tt{R_{eq_2}\:=\:\dfrac{4}{2}\Omega\:}$

➠ $\bf{R_{eq_2}\:=\:2\Omega\:}$

Case - 3 :

✯ $\rm{R_{eq_2}}$ & 4th resistor are connected in series.

Thus,

➠ $\tt{R_{eq_3}\:=\:2\Omega\:+\:2\Omega\:}$

➠ $\bf{R_{eq_3}\:=\:4\Omega\:}$

Case - 4 :

✯ $\rm{R_{eq_3}}$ & 5th resistor are connected in parallel.

Thus,

➠ $\tt{\dfrac{1}{R_{eq_4}}\:=\:\dfrac{1}{4\Omega}\:+\:\dfrac{1}{4\Omega}\:}$

➠ $\tt{\dfrac{1}{R_{eq_4}}\:=\:\dfrac{2}{4}\Omega\:}$

➠ $\tt{R_{eq_4}\:=\:\dfrac{4}{2}\Omega\:}$

➠ $\bf{R_{eq_4}\:=\:2\Omega\:}$

Case - 5 :

✯ $\rm{R_{eq_4}}$ & 6th resistor are connected in series.

Thus,

➠ $\tt{R_{eq_5}\:=\:2\Omega\:+\:2\Omega\:}$

➠ $\bf{R_{eq_5}\:=\:4\Omega\:}$

Case - 6 :

✯ $\rm{R_{eq_5}}$ & 7th resistor are connected in parallel.

Thus,

➠ $\tt{\dfrac{1}{R_{eq_6}}\:=\:\dfrac{1}{4\Omega}\:+\:\dfrac{1}{4\Omega}\:}$

➠ $\tt{\dfrac{1}{R_{eq_6}}\:=\:\dfrac{2}{4}\Omega\:}$

➠ $\tt{R_{eq_6}\:=\:\dfrac{4}{2}\Omega\:}$

➠ $\bf\pink{R_{eq_6}\:=\:2\Omega\:}$

∴ The equivalent resistance between P & Q is 2 Ω.