The circuit shown above consists of a single battery, whose emf is 1.6 V, and th

Question

The circuit shown above consists of a single battery, whose emf is 1.6 V, and three wires made of the same material, but having different cross-sectional areas. Each thick wire has cross-sectional area 1.2e-6 m2, and is 25 cm long. The thin wire has cross-sectional area 6.2e-8 m2, and is 3.7 cm long. In this metal, the electron mobility is 4e-4 (m/s)/(V/m), and there are 8e+28 mobile electrons/m3.

Use the appropriate equation(s), plus the equation relating electron current to electric field, to solve for the factor that goes in the blank below:
EF = ___ * ED
Use the appropriate equation(s) to calculate the magnitude of ED
ED = ____ V/m
Use the appropriate equation(s) to calculate the electron current at location D in the steady state:
iD = _____ electrons/s A

The circuit shown above consists of a single battery, whose emf is 1.6 V, and three wires made of the same material, but having different cross-sectional areas. Each thick wire has cross-sectional area 1.2e-6 m2, and is 25 cm long. The thin wire has cross-sectional area 6.2e-8 m2, and is 3.7 cm long. In this metal, the electron mobility is 4e-4 (m/s)/(V/m), and there are 8e+28 mobile electrons/m3. Use the appropriate equation(s), plus the equation relating electron current to electric field, to solve for the factor that goes in the blank below: EF = * ED Use the appropriate equation(s) to calculate the magnitude of ED ED = V/m Use the appropriate equation(s) to calculate the electron current at location D in the steady state: iD = electrons/s A

Explanation / Answer

You cannot use E = emf/L here because the circuit is not constant! To that, you know that:
i = n*a*M*E
i = electron current
n = electron density
a = area
M = electron mobility
E = electric field
For the thin and thick wire, the electron current for both of them is constant. So you can put the two together!
nA(thick)ME(thick) = nA(thin)ME(thin)
A(thick)E(thick) = A(thin)E(thin), n and M are constants, too. :D
E(thick) = [A(thin)/A(thick)]E(thin)
A(thin) = 6.2 x 10-8 m2
A(thick) = 1.2 x 10-6 m2
Hence

E(thick) = 0.052E(thin) [Answer to the first question]

You have the equation:
0 = 1.6 - EF*0.25 - ED*0.037 - EF*0.25
0 = 1.6 - EF*0.5 - ED*0.037
E(thick) = 0.052E(thin)
So, EF = E(thick) and ED = E(thin) from the diagram. Plug it in.
0 = 1.6 - 0.052*0.5E(thin) - E(thin)*0.037
0 = 1.6 - 2(0.25)(0.052)E(thin)
0.0628E(thin) = 1.6
E(thin) = 25.5 V/M [Answer to the second question]

You have i=naME to find i. So plug that in.
M = 4 x 10-4
n = 8 x 1028
A(thin) = 6.2 x 10-8
E = 25.5 V/m
i = (4 x 10-4)(8 x 1028)(6.2 x 10-8)(25.5) = 5.1 x 1019 electrons / second [Answer to the last one]

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