The circuit shown above consists of a single battery, whose emf is 1.8 V, and th

Question

The circuit shown above consists of a single battery, whose emf is 1.8 V, and three wires made of the same material, but having different cross-sectional areas. Each thick wire has cross- sectional area 1.7e-6 m2, and is 17 cm long. The thin wire has cross-sectional area 6.5e-8 m2, and is 6.1 cm long. In this metal, the electron mobility is se-4 (m/sy/(v/m), and there are 7e+28 mobile electrons/m3 Which of the statements about the circuit in the steady state are true? At location B the electric field points toward the top of the page. The electron current at location D is the same as the electron current at location F The magnitude of the electric field at locations D and F is the same. The magnitude of the electric field at locations F and C is the same The symbol Ep represents the magnitude of the electric field at location F, and the symbol Ep represents the magnitude of the electric field at location D. Which of the following equations is a correct energy conservation (loop) equation for this dircuit, following a path that starts at the negative end of the battery and goes counter-dlockwise? . 0. +1.8V. Er"0.17m·ED-0.06 im.Ep-0.17m 0 = . 1.8V + Er·0.1 7m + ED-0.06 im + Er-0.17m ü 0-1.EV + E,-0.17m + ED·0,00 1m + Er"0.1 7m

Explanation / Answer

a)

from the equation,

0=1.8-EF*0.17-Ed*0.061-EF*0.17

0=1.8-0.038*0.17-Ed*0.061-0.038*0.17

===> Ed=29.3 v

b)

use,

id=n*A*u*Ed

id=7*10^28*6.5*10^-8*5*10^-4*29.3

id=6.66*10^19 electron/sec

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