A 3-V battery is connected through a switch to two identical resistors and an id
ID: 1909891 • Letter: A
Question
A 3-V battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors has a resistance of 141 ?, and the inductor has an inductance of 3 H. The switch is initially open. (Assume positive current runs downward, from top to bottom.)
(d) After a long time (> 10 s), the switch is opened again. Immediately after the switch is opened, what is the current in resistor R1 and in resistor R2? i1=[] i2 = []
(e) At 50 ms after the switch is opened, what is the current in resistor R1 and in resistor R2? i1 = [] i2 = []
(f) At 500 ms after the switch is opened, what is the current in resistor R1 and in resistor R2? i1 = [] i2 = []
A 3-V battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors has a resistance of 141 ?, and the inductor has an inductance of 3 H. The switch is initially open. (Assume positive current runs downward, from top to bottom.) (d) After a long time (> 10 s), the switch is opened again. Immediately after the switch is opened, what is the current in resistor R1 and in resistor R2? i1=[] i2 = [] (e) At 50 ms after the switch is opened, what is the current in resistor R1 and in resistor R2? i1 = [] i2 = [] (f) At 500 ms after the switch is opened, what is the current in resistor R1 and in resistor R2? i1 = [] i2 = []Explanation / Answer
d) assume that internal resistance is zero. so,there is a circuit symbol for an ideal voltage generator which can be used to give a precise indication of such a meaning. The symbol is simply a circle with the direction and magnitude of the emf drawn beside it. e) current in R1 = 3/(141 + rb) = 21.2mA (if rb is taken to be 0) The current (I) in the inductor is given by I = (V/R)*(1 - e^(-t*R/L)) where R = 141 after 50ms, I = 19.2mA f) current in R1 = 3/141 = 21.2 mA
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