The switch in the figure below is open for a very long time and then closed at t

Question

The switch in the figure below is open for a very long time and then closed at

t = 0.

(Assume

V = 6.0 V

and

R2 = 580 .)

(a) What is the current through R1 and through R2 immediately after the switch is closed?

(b) What is the voltage across the inductor immediately after the switch is closed?
V

(c) After the switch is closed for a very long time, what is the current through R1, the current through R2, and the voltage across L?

(d) The switch is now opened at

t = topen.

What are the magnitude and direction of the current through R1 immediately after topen?

(e) What is the voltage across L just after topen?
V

Explanation / Answer

a)

immediately after switch is closed , the inductor behaves as a open circuit

hence no current flows through R2 , so i2 = 0

henc R1 and R2 are in parallel with each other and the battery

i1 = current in R1 = V/R1 = 6/800 = 0.0075 = 7.5 mA

b)

Voltage across inductor = VL = V = 6 V

c)

after long time switch is closed , the inductor behaves as a short circuit

so VL = Voltage across inductor = 0

henc R1 and R2 are in parallel with each other and the battery

i1 = current in R1 = V/R1 = 6/800 = 0.0075 = 7.5 mA

i2 = current in R2 = V/R2 = 6/580 = 0.0103 = 10.3 mA

d)

after switch is opened , VL = voltage across inductor = Voltage of battery = 6

i = current in R1 = VL / (R1 + R2) = 6 / (800 + 580) = 0.00435 A

e)

after switch is opened , VL = voltage across inductor = Voltage of battery = 6

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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