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Individual electrical circuit with battery cables and 2 thick and thin wire?

The circuit http://www.webassign.net/mi3/19-072-ProblemThinSection.jpg shown above consists of a single battery whose emf is 1.4 V, and three cables of the same material but with different cross-sectional areas. Each wire thickness has sectional area 1.4e-6 m2, and is 21 cm long. The thin wire is 6.4e-8 m2 cross-sectional area, and is 7.5 cm long. In this metal, the electron mobility is 6e-4 (m / s) / (V / m), and there are 28 mobile 8e electrons/m3 -. 0 = 0.21 m 1.4 V * EF – ED * 0.075m – 0.21 m EF * Use the appropriate equation (s) plus the equation of the electron current on the electric field, to solve for the factor that goes in the space below: EF =? * Use the appropriate equation ED (S) to calculate the magnitude of ED =? Use the appropriate equation (s) to calculate the electron current in the position D in the steady state already tried, and I can not. E = V / L does not work here.

Greetings again, deadfeesh! I just want to know I was looking through my physics book for you, and I learned how. I'm surprised you posted this same question like 5 times – you really must have been confused! To pay for all the problems chemistry that helped me with, I hope this helps you, too! You may not use E = emf / L here because the circuit is not constant! For that, you know that: i = behalf of electrons i = n the current density of electrons = M = area electron = E = electrical mobility and thickness field for thin wire, the electron current for both is constant. So you can put the two together! NA (thick) ME (thickness) = nA (thin) ME (thin) A (thickness) E (thickness) = A (thin) E (fine) N and M are constant, too. : D, E (thickness) = [A (fine) / A (thick)] E (thin) A (thin) = 6.4E-8 m ^ 2 A (thickness) = 1.4E-6 m ^ 2 E (thickness ) = 0.046E (fine) [Reply to Question] You have the equation: 0 = 1.4 – EF * 0.21 – * 0075 ED – EF * 0.21 E (thickness) = 0.046E (thin) =. For Therefore, EF E (thickness) and ED = E (fine) in the diagram. That connect 0 = 1.4 – 2 (0.21) (0.046) E (thin) E (thin) = 14.8 V / M [Reply to the second concerned] you have i = name to find i. So plug in M = 6E-4 n = 8E28 A (thin) = 6.4E-8, E = 14.8 i = (6E-4) (8E28) (6.4E-8) (14.8) = 4.5E19 electrons / second [Response to the latest] Hope this helps, deadfish!

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