A wire having length 2.0 m diameter 1.0 mm and resistivity `1.963 xx 10^(-8) Ω` m is connected in series with a battery of emf 3V and internal resistance I`Ω`. Calculate the resistance of the wire and current in the circuit.

Explanation

Here ` =(rho l)/(A) = ( rho l)/( pi r^2)=( 1.963 xx 10^(-8) xx 2) /( 3.14 xx (0.5 xx 10^(-3))^2)`
`R= (3.926 xx 10^(-8) )/( 0.785 xx 10^(-6)) `
` R= 5.00 xx 10^(-2) = 0.05 Omega `
current in the circuit is
`I=(E )/(R+r) =(3)/(0.05 +1)=(3)/(1.05 ) = 2.857 A `