A conducting wire of length T, area of crosssection A and electric resistivity ρ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be:

Question

A conducting wire of length T, area of crosssection A and electric resistivity ρ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be: (A) (1/4) ( VA /ρ l) (B) (3/4) ( VA /ρ l) (C) (1/4) (ρ l/ VA ) (D) 4 ( VA /ρ l)

Tardigrade Admin · Accepted Answer

As per the question <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/371672536fddd9a0c1f3e1902d3ed0d0-.png" /> Resistance =(ρ(2 l)/( A / 2))=(4 ρ l/ A ) ⇒ Current =( V / R )=( VA /4 ρ l)

$\frac{1}{4} \frac{V A}{ρl}$

$\frac{3}{4} \frac{V A}{ρl}$

$\frac{1}{4} \frac{ρl}{V A}$

$4 \frac{V A}{ρl}$

As per the question

Resistance $= \frac{ρ ( 2 l )}{( A /2 )} = \frac{4 ρl}{A}$
$\Rightarrow$ Current $= \frac{V}{R} = \frac{V A}{4 ρl}$

41​ρlVA​

43​ρlVA​

41​VAρl​

4ρlVA​

As per the question Resistance =(A/2)ρ(2l)​=A4ρl​ ⇒ Current =RV​=4ρlVA​

$\frac{1}{4} \frac{V A}{ρl}$

$\frac{3}{4} \frac{V A}{ρl}$

$\frac{1}{4} \frac{ρl}{V A}$

$4 \frac{V A}{ρl}$

As per the question

Resistance $= \frac{ρ ( 2 l )}{( A /2 )} = \frac{4 ρl}{A}$
$\Rightarrow$ Current $= \frac{V}{R} = \frac{V A}{4 ρl}$