Answers a, b, c and f are correct. I am looking for the solutions for part d and

Question

Answers a, b, c and f are correct. I am looking for the solutions for part d and e. Please show steps.

f is R1=0 and R2=0

A 9.00-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 0297 , and the inductor has an inductance of8.90 H. The switch is initially open Ri a) Immediately after the switch is closed, what is the current in resistor R1 and in resistor R2? A0303 b) At 50.0 ms after the switch is closed, what is the current in resistor R1 and in resistor R2?

Explanation / Answer

Voltage of battery = 9   volt

R1 = R2 = 297   ohm

Inductance L = 8.90 H

(d.) When the switch is open after a very long time the current in R1 and R2 will be same as the current in R2 just before the switch is open .
This will happen due to the inertia produced by the inductor.

Current in R2 just before the switch is open = (8/297) A = 1/33 A = 0.0303 A

Hence immediately after switched is ope , current in R1 = Current in R2 = 0.0303 A .

(e.) At 50 ms

Discharging formula of inductor

          I = I0 *e(-tReq / L )                   where Req = R1 + R2 = 594 ohm. and L = 8.90 H

or                I = 0.0303*e(-50*594*0.001/8.90 )                  

or    I   = 0.0303*0.0355 = 0.00108 A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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