Inductive Circuit Analysis
A 400-turn solenoid is 25-cm long and 3.0-cm in diameter.
(a) What is the inductance of the solenoid?
(b) The solenoid is then connected a circuit as shown in the figure, where e0=24.0V, R1=6.0ohm, R2=4.0ohm, R3=8.0ohm. Find current I1, I2 and I3 immediately after the switch is first closed.
(c) Find I1, I2 and I3 after the switch is closed a long time.
(d) Finally, the switch is turned off, what is the maximum induced emf on the solenoid and the current changing rate dl/dt immediately after the switch is off?
See attached file for diagrams.
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See the attachment.
Solenoid no. of turns N = 400
Length L = 25 cm or 0.25 m
Diameter D = 3 cm or 0.03 m
Area of cross section A = Π(d/2)2 = Π(0.03/2)2 = 7.07 x 10-4 m2
Inductance L = (μ0 x N2 x A)/L = (4Π x 10-7) x (400)2 x (7.07 x 10-4)/0.25 = 5.683 x 10-4 H
Immediately after the switch is closed, the current I3 in R3, L branch is zero. This is because, due to the presence of the inductance, the current rises gradually starting with zero value. Hence, immediately upon closure of the switch, I1 = I2 = E0/(R1+R2) = 24/10 = ...
Solution Summary
The solution involves an electric circuit problem. The circuit consists of a combination of three resistors and an inductor along with a battery and a switch. The solution uses the determination of various currents in the circuit immediately after the switch is closed and also immediately after the switch is opened.