A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire t

Question

A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire to the end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. The wire is at room temperature, so the resistivities are as given in Table 27.2. A potential difference of 5.0 V is maintained between the ends of the 2.0 m composite wire.

a.) What is the current in the copper section? b.) What is the current in the silver section?

b.) What is the magnitude of E in the copper section?

c.) What is the magnitude of E in the silver section?

d.) What is the potential difference between the ends of the silver section of the wire?

Explanation / Answer

Area

A=pi*d^2/4 =pi*(0.6*10^-3)^2/4

A=2.83*10^-7 m^2

Resistance of silver wire

Rs=psLs/A=(1.47*10-8)*1.2/(2.83*10-7) =0.0624 ohms

Resistance of copper wire

Rc=pcLc/A=(1.72*10-8)*0.8/(2.83*10-7) =0.0487 ohms

Total resistance of wire

R=Rs+Rc=0.111 ohms

current flowing in the wire

I=V/R =5/0.11

I=45 A

a)

Current in silver section

Is=45 A

b)

Current in coper section

Ic=45 A

c)

Current density

J=I/A =45/(2.83*10^-7)

J=1.59*108 A/m2

Electric field of Copper

Ec=Jpc=(1.59*10^8)(1.72*10^-8)

Ec=2.73 V/m

d)

Electric field of silver section

Es=Jps=(1.59*10^8)(1.47*10^-8)

Ec=2.34 V/m

e)

Potential difference across both ends of silver wire is

Vs=IRs=45*0.0624

Vs=2.81 Volts

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