Question

When a potential difference of 2V applied across the end of a wire of 5 m length , a current of 1 A is found to flow through it .calculate (¡) the resistance per unit length of the wire ,(¡¡) the resistance of 2 m of wire , (¡¡¡) the resistance accros the ends of the wire it is doubled on itself.

Answer 1

4 resistances can be formed by series and parallel combination of all 3 resistances.

Consider 3 resistances each of value R.

1. When all three are connnected in a seriescombination the equivalent resistance is 3R.

2. When any two of them are connected inparallel and one in series with them the equivalent resistance will be

(R×R/(R+R))+R=R/2+R=3/2R

3. When any two of them are connected inseries and one is connected in parallel then the effective resistance will be

(2R×R/(2R+R))=2/3R

4. When all three resistances are connected inparallel then the effective resistance will be

From point no.2 we can take the resistance of two resistances in parallel which is R/2

Hence the equivalent of three parallel will be R/2×R/(R/2+R)=R/3

Two more resistances can be obtained by putting any two in series(2R) and any two in parallel(R/2).

Answer 2

Here,

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) × 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 Ω