Question

A solenoid having an inductance of 3.3 ? is connected in series with a 4.7 Ω resistor. If a
10.0 ? battery is connected across the pair
Please Answer these 3 following questions.

(a) Find the equilibrium current in the circuit.
(b) How long will it take for the current to reach 80% of its final value?
(c) How much energy is dissipated by the resistor during the first 3 seconds?

Answer 1

The circuit equation by K.V.L can be given as-

V - iR - Ldi/dt = 0

=> Ldi/dt = V - iR

=> di/(V-iR) = dt/L

=> di/(V-iR) = dt/L

=> -1/R x ln(V-iR) = t/L + c

Now for constant c we know that inductor initially opposes the flow of current so at t= 0 we have I =0,

=> c = -1/R x ln(V)

=> 1/R x ln{V/(V-iR)} = t/L

=> ln{V/(V-iR)} = tR/L

=> V/(V-iR) = etR/L

=> i = V/R x (1-e-tR/L)

So we have at t =

=> i = V/R , This is the equilibrium current in circuit

B) Now for i = 80/100 x V/R = 0.8V/R

we have 0.8V/R = V/R x (1-e-tR/L)

=> e-tR/L = 0.2

=> -tR/L = ln(0.2)

=> t = -L/R x ln(0.2)

=> t = 1.61 s

c) Now power consumed by resistor is given as,

P = I2R

=>. P = {V/R x (1-e-tR/L)}2 R

=> P = V2/R x (1 - 2e-tR/L + e-2tR/L)

Now we have E = P.dt

=> E = V2/R x (1 - 2e-tR/L + e-2tR/L)dt

=> E = V2/R x [t + 2L/R x e-tR/L - L/2R x e-2tR/L]03

=>E = 41.8366 J