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

Explanation:

(i)

Let Resistance of a wire = R Ω

When cut into 3 equal parts

Resistance of each part = R*Ω/3

When connected in Parallel, net resistance Rp:

1/ Rp = 1/ R/3 + 1/ R/3 + 1/ R/3

= 9/R

Rp = R/9

and Resistivity will remain same.

(ii) Current through each part will be triple the previous and total current in the circuit will be 9 times.

Answer 2

Answer:-

$\pink{\bigstar}$ The resistance of the wire will become $\large\leadsto\boxed{\rm\purple{\dfrac{R}{9} \: \Omega}}$

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$\pink{\bigstar}$ The resistivity of the wire will $\large\leadsto\boxed{\rm\purple{remain \: same}}$

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

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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,

$\pink{\bigstar}$ $\underline{\boxed{\bf\green{\dfrac{1}{R_{eq.}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}}}$

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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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★ $\large{\underline{\underline{\bf\red{R_{eq.} = \dfrac{R}{9} \: \Omega}}}}$

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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)

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• Applying Ohm's Law:-

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$\pink{\bigstar}$ $\underline{\boxed{\bf\green{V = I \times R}}}$

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➪ $\sf I = \dfrac{V}{R}$

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➪ $\sf I = \dfrac{V}{\frac{R}{9}}$

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➪ $\sf I = \dfrac{9 \times V}{R}$

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★ $\large{\underline{\underline{\bf\red{I = 9 \times I}}}}$

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Therefore, the current increases 9 times the initial current.