Question

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Answer 1

☀️ Given that, two metallic wires A and B (of the same material) are connected in parallel.

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☀️ Wire A has length l and radius r.

☀️ Wire B has length 3l and radius 3r.

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☀️ We should find the ratio of the total resistance in parallel combination to the resistance of wire A.

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☀️ Finding the resistance of each wire :

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☀️ For wire A :

• Length = l
• Radius = r

$\sf\qquad:\implies~ R_{(A)}=Wire~A$

$\sf\qquad:\implies~{R_{(A)}=\rho}\dfrac{l}{A}$

$\sf\qquad:\implies~{ R_{(A)}=\rho}\dfrac{l}{\pi r^2}$

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☀️ For wire B :

• Length = 3l
• Breadth = 3r

$\sf\qquad:\implies~R_{(B)}=Wire~ B$

$\sf\qquad:\implies~R_{(B)}=\rho\dfrac{l}{A}$

$\sf\qquad:\implies~R_{(B)}=\rho\dfrac{l}{\pi r^2}$

$\sf\qquad:\implies~R_{(B)}=\rho\dfrac{3l}{\pi (3r)^2}$

$\sf\qquad:\implies~R_{(B)}=\rho\dfrac{3l}{\pi 9r^2}$

$\sf\qquad:\implies~R_{(B)}=\rho\dfrac{l}{3\pi r^2}$

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☀️ Finding the resistance of the parallel combination :

$\sf \qquad:\implies~ \dfrac{1}{R_{(P)}}=\dfrac{1}{R_{(A)}}+\dfrac{1}{R_{(B)}}$

$\sf\qquad:\implies{\dfrac{1}{ ~R_{(P)}}=}\dfrac{\sf 1}{\bigg(\dfrac{\sf{\rho }l}{\pi r^2}\bigg)}+\dfrac{\sf 1}{\bigg(\dfrac{{\sf{\rho}} l}{3\pi r^2}\bigg)}$

$\qquad:\implies~ \sf{\dfrac{1}{R_{(P)}}=\dfrac{\pi r^2}{\rho l}+\dfrac{3\pi r^2}{\rho l}}$

$\qquad:\implies~ \sf\dfrac{1}{R_{(P)}}=\dfrac{4\pi r^2}{\rho l}$

$\qquad:\implies~\sf R_{(P)}=\rho\dfrac{l}{4\pi r^2}$

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☀️ Finally, finding the ratio of the total resistance in parallel combination to the resistance of wire A :

$\qquad\sf :\implies~ R_{(P)}:R_{(A)}$

$\qquad\sf:\implies~\dfrac{R_{(P)}}{R_{(A)}}=\dfrac{\bigg(\dfrac{\rho l}{4\pi r^2} \bigg)}{\bigg(\dfrac{\rho l}{\pi r^2}\bigg)}$

$\sf\qquad :\implies~\dfrac{\rho l}{4\pi r^2}\times\dfrac{\pi r^2}{\rho l}$

$\sf{\qquad:\implies ~\sf \dfrac{R_{(P)}}{R_{(A)}}=\dfrac{1}{4}}$

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☀️ Therefore, the ratio of the total resistance in parallel combination to the resistance of wire A is 1 : 4.

Answer 2

Answer:

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