Question

A wire of 15 resistance is gradually stretched to double its original length.
It is then cut into two equal parts. These parts are then connected in parallel
across a 3 V battery. Find the current drawn from the battery​

Answer 1

Answer:

Explanation:

When the resistance of wire R=60Ω is cut into two equal parts, the length becomes half but area remains the same. Hence resistance of each part is R1= R/2 = 60/2 = 30Ω.. Hope this helps u!

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Answer 2

Answer:

Current in the circuit is 0.4 Ampere.

Explanation:

R = ∫ l / A

where:

R - resistance

∫ - resistivity

l - length

A - Area

∫ and A are constants here so don't bother about them.

R ∝ l

Case 1

When the length is doubled, new length ( l') = 2 l

Original resistance be R

New resistance be R'

Ratio between R and R' =

$\frac{R}{R'} = \frac{l}{l'}$

$\frac{15}{R'} = \frac{l}{2l}$

R' = 30 Resistance

Case 2

Wire is cut into 2 equal parts.

Resistance for each part:

Initial length (l')= 2l

Final length (l'')= l

$\frac{R'}{R''} = \frac{l'}{l''}$

$\frac{30}{R''} = \frac{2l}{l}$

Resistance for each part will be 15 N

Case 3

Wires are connected in parallel method.

Total resistance of the circuit be R.T

$\frac{1}{R.T} = \frac{1}{R1} + \frac{1}{R2}$

R.T = 7.5 Resistance

.

.

Total current in the circuit (I) :

I = V/R

 = 3/7.5

 = 0.4 Ampere

Conclusion: When u cut a wire in half, its resistance is also halved.

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