Question

a wire whose resistance is 80 ohm is cut into three pieces of equal lengths which are then arranged in parallel. calculate the resistance of the combination​

Answer 1

Answer:

8.99 ohm is the resistance of give combination

Answer 2

GivEn:

• A wire whose resistance is 80 ohm is cut into three pieces of equal lengths which are then arranged in parallel.

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To find:

• Resistance of the combination.

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SoluTion:

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${\underline{\bf{\bigstar\;As\;per\;given\;question\;:}}}$

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A long Cylindrical having resistance 80 ohm cut into three equal pieces.

Therefore, The resistance of each piece is ${}^{\text80}\!/{}_{\text{3}}\;\sf \Omega$

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All three pieces of wire are connected in parallel.

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Expression for equivalent resistance in parallel combination is,

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$\star\;{\boxed{\sf{\purple{ \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}}}}$

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Here, $\sf R_1 , R_2 , R_3 = {}^{\text80}\!/{}_{\text{3}}\; \Omega$

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$\;\;\;\;\;\small\sf \underline{Putting\;values\;:}$

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$:\implies\sf \dfrac{1}{R_p} = \dfrac{1}{ \frac{80}{3}} + \dfrac{1}{ \frac{80}{3}} + \dfrac{1}{ \frac{80}{3}}$

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$:\implies\sf \dfrac{1}{R_p} = \dfrac{3}{80} + \dfrac{3}{80} + \dfrac{3}{80}$

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$:\implies\sf \dfrac{1}{R_p} = \dfrac{3 + 3 + 3}{80}$

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$:\implies\sf \dfrac{1}{R_p} = \dfrac{9}{80}$

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$:\implies\sf R_p = \dfrac{80}{9}$

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$:\implies{\underline{\boxed{\bf{\pink{R_p = 8.89\;\Omega}}}}}\;\bigstar$

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$\therefore$ Hence, The equivalent resistance in parallel combination is $\bf 8.89\;\Omega$.