Question #6473

A wire of total length 5.00 m, consists of equal lengths of Cu followed by Al, both having the same diameter of 1.00 mm. A voltage difference of 89.0 V is placed across the composite wire. What is the current flowing in the wire? Use rho=1.72x10-8 ohm-m for the resistivity of copper and rho=2.82x10-8 ohm-m for the resistivity of aluminum.

Expert's answer

We can imagine two pieces of wire as two resistors - Aluminum and Cuprum. Then

the sum of this resistances will give us the necessary resistance

R.

R=R1+R1=rho(cu)xl(cu)/S(cu)+rho(al)xl(al)/S(al)={l(cu)=l(al)=l/2,

S(cu)=S(al)=pi*(d/2)^2}=(1,72+2,82) x 10^(-8)omh*m x 2.5m/(0.78 x

10^(-6)m)=0,1445Omh;

I=U/R=89V/0,1445Omh=615.86A

the sum of this resistances will give us the necessary resistance

R.

R=R1+R1=rho(cu)xl(cu)/S(cu)+rho(al)xl(al)/S(al)={l(cu)=l(al)=l/2,

S(cu)=S(al)=pi*(d/2)^2}=(1,72+2,82) x 10^(-8)omh*m x 2.5m/(0.78 x

10^(-6)m)=0,1445Omh;

I=U/R=89V/0,1445Omh=615.86A

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