Two resistors R 1 = 15.0 , and R 2 = 10.5 are connected with a 310 mH inductor,

Question

Two resistors R1 = 15.0 , and R2 = 10.5 are connected with a 310 mH inductor, a 12.0 V battery and a two-way switch as shown in the diagram below. At t = 0, the switch ab is closed.

(a) Determine the time constant for this circuit.
s

(b) Calculate the current in the two resistors and the inductor a long time after the switch is closed.

(c) What is the voltage across the two resistors and the inductor a long time after the switch is closed?

Now the switch ab is opened and ac is closed.

(d) Determine the time constant for this circuit after ac is closed. (Enter your answer to at least four decimal places.)
s

(e) What is the current in the inductor at t = 0.008 s after ac is closed?
A

(f) What is the voltage across the two resistors and the inductor at t = 0.008 s after ac is closed?

Explanation / Answer

firstly R1 and R2 are in series so R = R1+R2

R =15+10.5=25.5

now
a) time constant T = L/R
T = 310*10-3/25.5
T = 12.15*10-3 s
T = 12.15 ms
b)
After a long time the inductor will appear to be a short circuit, and the maximum current is
I = V/R
I = 12/25.5
I = 0.470 A

e)
i = Io*(1 - e-t/T)
i = 0.470*(1 - e-0.08/12.15)
i = 3.084*10-3A

i = 3.084 mA

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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