Question

An inductor with an inductance of 3.50 H and a resistance of 8.00 Ω is connected to the terminals of a battery with an emf of 4.00 V and negligible internal resistance.a)
Just after the circuit is completed, at what rate is the battery supplying electrical energy to the circuit?Express your answer with the appropriate units.b)When the current has reached its final steady-state value, how much energy is stored in the inductor?Express your answer with the appropriate units.c)What is the rate at which electrical energy is being dissipated in the resistance of the inductor?Express your answer with the appropriate units.d)What is the rate at which the battery is supplying electrical energy to the circuit?

Answer 1

Answer:

a) Δt = 0.437 sec. b) E = 0.4375 J.

Explanation:

a)

First we find the current with V = IR

I = V/R

I = 4/8

I = 1/2.

Now, we find the "time" with the formula:

emf = L×Δi / Δt

Δt = L×Δi / emf

Δt = 3.5×0.5 / 4

Δt = 0.437 sec.

b)

We find the energy stored in the inductor with the formula:

E = 1/2×LI^2

E = 1/2×3.5×(0.5)^2

E = 0.4375 J.