A battery is connected in series with a 0.28-? resistor and an inductor, as show

Question

A battery is connected in series with a 0.28-? resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is 0.12 s, and the maximum current in the circuit is 8.2 A.

(a) Find the emf of the battery.
V

(b) Find the inductance of the circuit.
mH

(c) Find the current in the circuit after one time constant has elapsed.
A

(d) Find the voltage across the resistor after one time constant has elapsed.
V

(e) Find the voltage across the inductor after one time constant has elapsed.
V

Explanation / Answer

Please ask if you have any doubt.I will help you.

a)The maximum current gives Vbattery = 8.2A(0.28) = 2.296V

Vb = 2.296

b) time constant = 0.12s = L/R gives L = (0.12)(0.28)

L = 33.6mH

c) The current increases to 63% of maximum valie after one time constant.

i() = 8.2(0.63) = 5.17A

d) voltage = i(R) = (5.17A)(.28) = 1.448V

e) VL = Vb - V(R) = 2.296 - 1.448 = 0.85V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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