Question

When a potential difference is applied to the piece of metal wire, a current of 4.1 mA flows through it. if this wire is replaced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire?

Assume that the lengths of the potential difference across both wires are the same. ( the resistivity of the original metal is 1.68x10^-2 ohm . m, and the resistivity of silver is 1.6x10^-8 ohm . m )

Answer 1

Let us assume,

Potential difference across each wire = V,

Length of each wire = L.

Given :

For the original wire,

Current = I_o = 4.1 mA,

Resistance = R = _oL / A,

where,

_o = Resistivity of this wire = 0.0168 ohm . m,

A = its area of cross - section = r², r being the radius of the wire.

For the silver wire,

Current = I_s = unknown,

Resistance = R^/ = _sL / A^/,

where,

_s = Resistivity of the silver wire = 1.6 x 10^-8 ohm . m,

A^/ = its area of cross - section = x ( 2r )², 2r being the radius of this wire.

Hence,

V = I_oR = I_sR^/

or, I_s = I_oR / R^/

or, I_s = ( I_ooL / A ) / ( _sL / A^/ ) = ( I_ooL / r² ) / [ _sL / x ( 2r )² ]

or, I_s = 4I_oo / _s.

Putting values :

I_s = ( 4 x 4.1 mA x 0.0168 ) / ( 1.6 x 10^-8 )

or, I_s = 1.72 x 10⁷ mA = 1.72 x 10⁴ A.

Hence, current in the silver wire is : 1.72 x 10⁴ A.