Question

When a potential difference of 2 V is applied across the ends of a wire of 5 m length, a current of 1A is found to flow through it. Calculate: (i) the resistance per unit length of the wire (ii) the resistance of 2 m length of this wire (iii) the resistance across the ends of the wire if it is doubled on itself.

Answer 1

Voltage, V = 2 V

Length of wire, L = 5 m

Current, I = 1 A

Now,

Resistance of the wire is, R = V/I = 2/1 = 2 Ω

1.

Resistance per unit length is = 2/L = 2/5 = 0.4 Ω/m

2.

Resistance of 2 m long wire will be = (0.4) x 2 = 0.8 Ω

3.

Let the resistivity of the material of the wire be q. Its original area of cross-section be A.

So, R = qL/A

= R = qL2/(AL)

= R = qL2/v

‘v’ is the volume of the wire.

When the length is doubled, its volume will remain constant. Also resistivity of the wire is constant.

So, new resistance is,

R/ = q(2L)2/v = 4[qL2/v]

=> R/ = 4R = 4 × 2 = 8

Answer 2

Answer: (i) `R = (V)/(I) = (2 V)/(1 A) = (2 V)/(1 A) = 2 Omega

Explanation:

(i) R=V/I=2V/1A=2V/1A=2Ω. Resistance//unit length =2Ω/5m=0.4Ω/m

(ii) Resistance of 2m length of the wire =0.4×2=0.8Ω

(iii) When the wire is doubled on itself, it will be equivalent to two resistances (each of value 1Ω) parallel to each other. The resultant resistance across the ends of this folded wire =1Ω/2=0.5Ω.

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