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00% H Tue Apr 4 23:06 a E Chrome File Edit View History Bookmarks People Window Help https://www.youtube.comwatc x VA Lab mo-inductance; Resonanc x C Home I Chegg.com f C Secure https:/ www.webassig web/Student/Assignment-Responses/last?dep 15435080 n.net/ a Apps E mur Jonsdottir o... User Dashboard Google Translate G Google is WebAssign LOG IN D Other Bookmarks -833 points UCFColPhysEML 0.Q.00 My Notes Two resistors R. 15.0 n, and R2 11.5 n are connected with a 330 mH inductor, a 12.0 V battery and a two-way switch as shown in the diagram below. Att 0, the switch ab is closed. (a) Determine the time constant for this circuit. (b) Calculate the current in the two resistors and the inductor a long time after the switch is closed R2 (c) What is the voltage across the two resistors and the inductor a long time after the switch is closed? Now the switch ab is opened and ac is closed (d) Determine the time constant for this circuit after acis closed. (Enter your answer to at least four decimal places (e) What is the current in the inductor at t 0.006 s after ac is closed? (f) What is the voltage across the two resistors and the inductor at t 0.006 s after ac is closed? Additional MaterialsExplanation / Answer
a)
Time constant = T = L/R1 = 0.330/15 = 0.022 sec
b)
R2 is not connected in the circuit when switch ab is closed , hence
i2 = current in R2 = 0
after long time , the inductor behaves as short circuit
hence using ohm's law
i1 = current in R1 = V/R1 = 12/15 = 0.8 A
iL = current ininductor = i1 = 0.8 A
c)
VR2 = i2 R2 = 0 (11.5) = 0 volts
VR1 = i1 R1 = 0.8 x 15 = 12 volts
after long time , inductor offer 0 resistance since it behaves as short circuit
VL = iL XL = 0
d)
T = time constant = L / (R1 + R2) = 0.33/(15 + 11.5) = 0.0125 sec
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