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 = 170 Ohm, the inductance is L = 2.80 H, and the emf of the battery is E = 31.0 V. At time t = 0 the switch is closed. What is the current through the circuit after the switch has been closed for t = 1.20 times 10^-2 s? What is the potential difference across the inductor after the switch has been closed for t = 1.20 times 10^-2 s? What is the power dissipation in the resistor at t = 1.20 times 10^-2 s? How much energy is stored in the inductor at t = 1.20 times 10^-2 s? How much work has the battery done from the time the switch was closed until t = 1.20 times 10^-2s? How much energy has been dissipated in the resistor from the time the switch was closed until t = 1.20 times 10^-2 s?

Explanation / Answer

formula for growth of current in LR circuit is

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

(a)

I= (31/170)*(1-e^-(1.2*10^-2*150/2.8))

i = 0.086A

(b)

(b)

V = E- IR = 31-0.086(170)

=16.38 V

(c)

P= i^2*R = 0.086^2*170 = 1.25W

(d)

U = 1/2 LI^2 = 1/2 ( 2.8) (0.086)^2 = 0.0103544 J

As per guide lines I worked first four parts kinedly post remining parts in the next question

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