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 m", and is 25 cm long. The thin wire has cross-sectional area 4.5e-8 my, and is 3.9 cm long. In this metal, the electron mobility is 4e-4 (m/s)/(V/m), and there are 8e+ 28 mobile electrons/m". Which of the following statements about the circuit in the steady state are true? The electron current at location D is the same as the electron current at location F. At location B the electric field points toward the top of the page. 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 circuit, following a path that starts at the positive end of the battery and goes clockwise? O 0 = +1.8V - EF*0.25m - Ep*0.039m - EF*0.25m 0 = -1.8V + EF*0.25m + ED*0.039m + EF*0.25m Do = -1.8V - E*0.25m - ED*0.039m - EF*0.25m Do = 1.8V + E,*0.25m + Ep*0.039m + EF*0.25m 1.8V = E,*0.25m O 1.8V = ED*0.039m The symbol is represents the electron current at location F, etc. Which of the following equations is a correct charge conservation (node) equation for this circuit? O iF tip = 0 O iF = 23ip O 2*iF = ip O iF = ip 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: D = electrons/s Additional Materials eBook