A 18.00-V battery is connected through a switch to two identical resistors and a

Question

A 18.00-V battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors a resistance of 200. Ohm, and the inductor has an inductance of 6,00 H. The switch is initially open. a) Immediately after the switch is closed, what is the current in resistor R1 and in resistor R2? b) At 40.0 ms after the switch is closed, what is the current in resistor R1 and in resistor R2? c) At 400. ms after the switch is closed, what is the current in resistor R1 and in resistor R2? d) After a long time (> 20.0 s), the switch is opened again. Immediately after the switch is opened, what is the current in resistor R1 and in resistor R2? e) At 40.0 ms after the switch is opened, what is the current in resistor R1 and in resistor R2? f) At 400. ms after the switch is opened, what is the current in resistor R1 and in resistor R2? Each small questions are 7 points.

Explanation / Answer

a) i = V/R1 =18/200 = 0.09 A answer

b) i1 = 0.09A

i2 = 0.09(1-e^(-0.040*200/6)) = 0.0663 A

C) i1 = 0.09 A

i2 = 0.09(1-e^(-0.40*200/6)) = 0.0899999 A

D) since current in inductor cant change abruptly,

i2 = 0.09 A

i1=0.09 A

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