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jtw2e
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Current in wire -- Please help
A 1.63-m length of wire is made by welding the end of a 100-cm long silver wire to the end of a 63-cm long copper wire. Each piece of wire is 1.1 mm in diameter. A potential difference of 5.0 mV is maintained between the ends of the 1.63-m composite wire.
ρsilver = 1.47e-8Ωm
ρcopper = 1.72e-8Ωm
What is the current in the silver section? (mA)
What is the current in the copper section? (mA)
What is the potential difference (voltage drop) between the ends of the silver section of wire? (mV)
What is the magnitude of the electric field in the silver? (N/C)
Part A:
I assumed we had to find the resistance in each part, sum them, then find the current. The current from each section should be the same, correct? The value I have below (148 milliAmps) is wrong according to the online homework system.
Rs = ρL/A
= [(1.47e-8Ωm)(1.0m)]/[π(.0011m/2)^2]
= 0.015468Ω
Rc = ρL/A
= [(1.72e-8Ωm)(.63m)]/[π(.0011m/2)^2]
= 0.018098Ω
Rtotal = Rs + Rc = 0.033567Ω
V = IR
I = V/R
= (0.005V)/(0.033567Ω)
= 0.14895 Amp
= 148.954 milliAmps
Homework Statement
A 1.63-m length of wire is made by welding the end of a 100-cm long silver wire to the end of a 63-cm long copper wire. Each piece of wire is 1.1 mm in diameter. A potential difference of 5.0 mV is maintained between the ends of the 1.63-m composite wire.
ρsilver = 1.47e-8Ωm
ρcopper = 1.72e-8Ωm
Homework Equations
What is the current in the silver section? (mA)
What is the current in the copper section? (mA)
What is the potential difference (voltage drop) between the ends of the silver section of wire? (mV)
What is the magnitude of the electric field in the silver? (N/C)
The Attempt at a Solution
Part A:
I assumed we had to find the resistance in each part, sum them, then find the current. The current from each section should be the same, correct? The value I have below (148 milliAmps) is wrong according to the online homework system.
Rs = ρL/A
= [(1.47e-8Ωm)(1.0m)]/[π(.0011m/2)^2]
= 0.015468Ω
Rc = ρL/A
= [(1.72e-8Ωm)(.63m)]/[π(.0011m/2)^2]
= 0.018098Ω
Rtotal = Rs + Rc = 0.033567Ω
V = IR
I = V/R
= (0.005V)/(0.033567Ω)
= 0.14895 Amp
= 148.954 milliAmps
Last edited: