A wire of uniform cross-section and length l has a resistance of 16 Ω. It is cut into four equal parts. Each part is stretched uniformly to length l and all the stretched parts are connected in parallel. Calculate the total resistance of the combination so formed. (Assume that the stretching of the wire does not cause any change in the density of the material.)

Question

A wire of uniform cross-section and length l has a resistance of 16 Ω. It is cut into four equal parts. Each part is stretched uniformly to length l and all the stretched parts are connected in parallel. Calculate the total resistance of the combination so formed. (Assume that the stretching of the wire does not cause any change in the density of the material.) (A) 10 Ω (B) 12 Ω (C) 16 Ω (D) 18 Ω

Tardigrade Admin · Accepted Answer

In stretching 4 times R becomes 42 time R prime=(R/4)=(16/4)=4 R prime prime=42 × 4=64 Each part has (1/R eq )=(1/64)+(1/64)+(1/64)+(1/64)=(4/64) R eq =16 Ω

$10 Ω$

$12 Ω$

$16 Ω$

$18 Ω$

In stretching 4 times $R$ becomes $4^{2}$ time
$R^{'} = \frac{R}{4} = \frac{16}{4} = 4$
$R^{''} = 4^{2} \times 4 = 64$
Each part has $\frac{1}{R _{e q}} = \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} = \frac{4}{64}$
$R_{e q} = 16Ω$

10Ω

12Ω

16Ω

18Ω

In stretching 4 times R becomes 42 time R′=4R​=416​=4 R′′=42×4=64 Each part has Req​1​=641​+641​+641​+641​=644​ Req​=16Ω

$10 Ω$

$12 Ω$

$16 Ω$

$18 Ω$

In stretching 4 times $R$ becomes $4^{2}$ time
$R^{'} = \frac{R}{4} = \frac{16}{4} = 4$
$R^{''} = 4^{2} \times 4 = 64$
Each part has $\frac{1}{R _{e q}} = \frac{1}{64} + \frac{1}{64} + \frac{1}{64} + \frac{1}{64} = \frac{4}{64}$
$R_{e q} = 16Ω$