A 12.00-V battery is connected through a snitch to two identical resistors and a
ID: 1517689 • Letter: A
Question
A 12.00-V battery is connected through a snitch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors has a resistance of 100. Ohm, and the inductor has an inductance of 4.00 H. The switch is initially open. Immediately after the switch is closed, what is the current in resistor R1 and in resistor R2 At 50.0 ms after the switch is closed, what is the current in resistor R1 and in resistor R2? At 500, ms after the switch is closed, what is the current in resistor R1 and in resistor R2?Explanation / Answer
Here ,
V = 12 V
L = 4 H
R = 100 Ohm
a) just after closing the switch
current in R1 = V/R1
current in R1 = 12/100
current in R1 = 0.12 A
as initially , as the inductor is open circuited
the current in the R2 is Zero.
b)
after 50 ms
as the current in R1 is constant ,
the current in R1 is 0.12 A
Time constant for circuit is
T = L/(R1 || R2)
T = 4/(100/2) = 8 ms
at t = 50 ms
current in R2 = 0.12 * (1 - e^(-50/50))
current in R2 = 0.0758 A
c) at t = 500 ms
as the current in R1 is constant ,
the current in R1 is 0.12 A
Time constant for circuit is
T = L/(R1 || R2)
T = 4/(100/2) = 8 ms
at t = 500 ms
current in R2 = 0.12 * (1 - e^(-500/50))
current in R2 = 0.119995 A
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