12. 2/4 points | Previous Answers YF13 29.P.049 My Notes Ask Your Teacher In the
ID: 1794745 • Letter: 1
Question
12. 2/4 points | Previous Answers YF13 29.P.049 My Notes Ask Your Teacher In the circuit shown in the figure below the capacitor has capacitance C 22 F and is initially charged to 10 V with the laity shown Theres sto R ha ressistance At time t 0 the switch is closed. The small circuit is not connected in any way to the large one. The wire of the small circuit has a resistance of 1.0 /m and contains 26 loops. The large circuit is a rectangle 2.0 m by 4.0 m, while the small one has dimensions a = 10.0 cm and b = 20.0 cm. The distance c is 5.0 cm. (The figure is not drawn to scale.) Both circuits are held stationary. Assume that only the wire nearest the small circuit produces an appreciable magnetic field through it. Ro (a) Find the current in the large circuit 280 s after s is closed. (b) Find the current in the small circuit 280 s after S is closed HAExplanation / Answer
at timt t = 0,
Io = Vmax/R
= 100/10
= 10 A
a) Time constant of the ckt, T = R*C
= 10*22*10^-6
= 220 micro s
at time t = 280 micro s,
I = Io*e^(-t/T)
= 10*e^(-280/220)
= 2.8 A <<<<<<------------------------Answer
b) magnetic flux through the loop = N*integral B*b*dl
= N*integral (mue*I/(2*pi*r))*b*dl
= N*mue*I*b/(2*pi)*ln((c+a)/c)
incued emf in the loop = N*mue*(dI/dt)*b/(2*pi)*ln((c+a)/c)
dI/dt = Io*e^(-t/T)*(1/T)
induced current = induced emf/R
= N*mue*Io*e^(-t/T)*(1/T)*b*ln((c+a)/c)/(2*pi*R)
= 26*4*pi*10^-7*10*e^(-280/220)*(1/(220*10^-6))*0.2*ln(15/5)/(2*pi*26*0.6*1)
= 932 mico A <<<<<<------------------------Answer
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