A 12.00-V battery is connected through a switch to two identical resistors and a

Question

A 12.00-V battery is connected through a switch to two identical resistors and an ideal inductor, as shown in the figure. Each of the resistors 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

Rtotal = R/2 = 100/2 = 50 ohm

(a)

current immediately after switch is closed is

i = V/Rt = 12/50 = 0.24 A

current is equal in both R1 and R2 = i/2 = 0.12 A

(b)

current after 50 ms

i' = i*e^(-t*R/L)

i' = 0.24*e^(-0.05*50 / 4)

i' = 0.24* 0.e^(-0.625)

i' = 0.128

so current in R1 and R2 = 0.128/2 = 0.064 A

(c)

current after 500 ms

i' = 0.24*e^(-0.5*50/4)

i' = 0.24*e^(-6.25)

i' = 0.46*10^(-3) A

current in R1 and R2 = 0.23*10^(-3) A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.