A 2.30 mH inductor is connected in series with a dc battery of negligible intern

Question

A 2.30 mH inductor is connected in series with a dc battery of negligible internal resistance, a 0.880 kOmega resistor, and an open switch. How long after the switch is closed will it take for the current in the circuit to reach half of its maximum value ? How long after the switch is closed will it take for the energy stored in the inductor to reach half of its maximum value? Please dont copy and paste an incorrect method of solving form another source. An please explain how to solve. Thank you.

Explanation / Answer

I = I0 e^(-t/tau)
tau = L/R
0.5= e^(-t*0.88E3/2.3E-3)
t=1.81E-6 s

E inductor = 1/2 L I^2
so to halve this I = I0/sqrt(2)
1/sqrt(2)=e^(-t*0.88E3/2.3E-3)
t=9.06 E-7 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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