A wire is cut into four pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was R , the resistance of the bundle will be:

Question

A wire is cut into four pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was R , the resistance of the bundle will be: (A)  (R/4)  (B)  (R/8)  (C)  (R/16)  (D)  (R/32)

Tardigrade Admin · Accepted Answer

For a wire of length l, area of cross-section A, and specific resistance ρ, the resistance is given by R=(ρ l/A) From above equation it is clear that resistance is directly proportional to length. When wire is cut into 4 pieces then resistance of each part is R'' propto (l/4) ⇒ R'' =(R/4) Also, equivalent resistance for parallel combination is (1/R')=(1/R1)+(1/R2)+(1/R3)+(1/R4) ∴ (1/R')=(4/R'')=(4 × 4/R) ⇒ R'=(R/16)

Q. A wire is cut into four pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was $R$ , the resistance of the bundle will be:

$\frac{R}{4}$

$\frac{R}{8}$

$\frac{R}{16}$

$\frac{R}{32}$

Q. A wire is cut into four pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was R , the resistance of the bundle will be:

4R​

8R​

16R​

32R​

Q. A wire is cut into four pieces, which are put together by sides to obtain one conductor. If the original resistance of wire was $R$ , the resistance of the bundle will be:

$\frac{R}{4}$

$\frac{R}{8}$

$\frac{R}{16}$

$\frac{R}{32}$