Question

A wire is cut into three equal parts and then connected in parallel with the same source. How will its.
(i) resistance and resistivity gets affected?
(ii) How would the total current and the current through the parts change?

Answer 1

Answer:

1)there will be a definite change in both 2) total may not flow frequently so the current may be very low

Answer 2

Answer:-

$\red{\bigstar}★ \sf The \: resistance \: of \: the \: wire \: will \: become \large\leadsto\boxed{\sf\purple{\dfrac{R}{9} \: \text{\O}mega}}$

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$\red{\bigstar}★ \sf \: The \: resistivity \: of \: the \: wire \: will \large\leadsto\boxed{\sf\purple{remain \: same}}$

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$\red{\bigstar}★ \sf The \: total \: current \: and \: the \: current \: through \: the \: parts \: will \large\leadsto\boxed{\sf\purple{increase \: 9 \: times}}$

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Solution:-

i)Let the initial resistance be 'R' when the wire in uncut.

Now,

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• According to the question:-

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The wire is cut into three equal parts and then connected in parallel with the same source.

Hence,

The new resistance will be 'R/3'.

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• Resistance:-

We know,

$\red{\bigstar}★ \underline{\boxed{\sf\pink{\dfrac{1}{R_{eq.}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}}}$

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$➪ \sf \dfrac{1}{R_{eq.}} = \dfrac{1}{\frac{R}{3}} + \dfrac{1}{\frac{R}{3}} + \dfrac{1}{\frac{R}{3}}$

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$➪ \sf \dfrac{1}{R_{eq.}} = \dfrac{3}{R} + \dfrac{3}{R} + \dfrac{3}{R}$

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$➪ \sf \dfrac{1}{R_{eq.}} = \dfrac{9}{R}$

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$★ \large{\underline{\underline{\sf\pink{R_{eq.} = \dfrac{R}{9} \: \text{\O}mega}}}}$

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• Resistivity:-

✯ Resistivity depends on the nature of the material. Therefore, no change in resistivity of the wire.

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ii)⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Applying Ohm's Law:-

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$\red{\bigstar}★ \underline{\boxed{\sf\pink{V = I \times R}}}$

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$➪ \sf I = \dfrac{V}{R}$

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$➪ \sf I = \dfrac{V}{\frac{R}{9}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$➪ \sf I = \dfrac{9 \times V}{R}$

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$★ \large{\underline{\underline{\sf\pink{I = 9 \times I}}}}$

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Therefore, the current increases 9 times the initial current.

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