The switch in the RL circuit shown is initially open. The resistance in the circ

Question

The switch in the RL circuit shown is initially open. The resistance in the circuit is R = 160 Ohm, the inductance is L - 3.40 H, and the emf of the battery is E = 50.0 V. At time t = 0 the switch is closed. What is the current through the circuit after the switch has been closed for t = 3.02 times 10^-2 s? What is the potential difference across the inductor after the switch has been closed for t = 3.02 times 10^-2 s? What is the power dissipation in the resistor at t = 3.02 times 10^-2 s? How much energy is stored in the inductor at t = 3.02 times 10^-2 s?

Explanation / Answer

The current in the LR circuit increases as

i = (E/R)*(1-e^-(tR/L))

part(a)

at t = 3.02*10^-2 s

i = (50/160)*(1-e^-(3.02*10^-2*160/3.4))

i = 0.24 A

===============

part (b)

V = E- i*R = 50 - (0.24*160) = 11.6 V

=============

part(c)

Power = i^2*R = 0.24^2*160 = 9.22 W

==============

part(d)

energy = (1/2)*L*i^2 = (1/2)*3.4*0.24^2 = 0.098 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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