A series LR circuit, with L=11.0 H and R= 25 Q. is connected to a 12.0 V batten

Question

A series LR circuit, with L=11.0 H and R= 25 Q. is connected to a 12.0 V batten and a series switch, initially open. At t=0 the switch is closed. What is the initial current in the circuit (just after t=0)? What is the current in the circuit after waiting a long time for equilibrium? 1 point A What is the energy stored in the inductor at this equilibrium? At a time.t_9. when the voltage across the inductor is 9.0 V. what is the current in the circuit? What is the rate of power dissipation in the resistor at this time. t_9?

Explanation / Answer

Given,

L = 11 H and R = 25 Ohm ; V = 12 V

1)Intial current just after the switch is closed will be zero.

2)Current after a long time will be:

I = V/R = 12/25 = 0.48 A

3)Energy stored is given by:

E = 1/2 L I2

E = 0.5 x 11 H x 0.48 x 0.48 = 1.27 J

4)We know that

V(L) = V(battery) x e-tR/L

9 = 12 x e-t x 25/11

e-t x 2.27 = 0.75

taking natural log both the sides:

-t x 2.27 = -0.288

t = 0.288/2.27 = 0.127 s

Hence, t = 0.127 s

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