A circuit is designed as shown below Use L 5 H R1 10 ohm R

A circuit is designed as shown below. Use L = 5 H, R_1 = 10 ohm, R_2 = 2.5 ohm and e = 100 V. The switch is flipped to position A. Determine the time required for the power delivered to the resistor to reach 75% of its maximum value. At that instant (which is the answer to part a), determine the flux through each turn of the inductor. Assume that the inductor has 100 turns. At that instant (which is the answer to part a), the switch is flipped to position B. Determine the voltage across the inductor 0.25s later. Verify energy conservation for the circuit after the switch has been flipped to position B. In other words, demonstrate that the energy initially stored by the inductor is all delivered to the resistors. Assume that R_1 is a light bulb. Qualitatively graph the current through the bulb as a function of time and describe how its brightness changes. Start from the instant you flipped the switch to A until the time when the bulb has completely gone out. (In a class demonstration, we saw that there was a delay for some bulbs to light since they had to heat up sufficiently. Please assume that those delays are negligible here.)

Solution

growth of curent in an LR circuit is given by

i = i₀ (1-exp(-Rt/L))    - i₀   is the peak current reached at infinity.-

i₀ = E/R = 100/10 = 10 amp

power in the resistor = i²R

when power is 75% of it maximum

(i/ i₀ )² = (1-exp(-Rt/L))² = 0.75

exp(-Rt/L) = 1 - 0.87 = 0.13

Rt/L = 2.04

t = 2.04 *5/10 = 1.02 s

i   = 0.87 *10 = 8.7 mp

Flux through the inductor = Li/N = 5*1.02/100 = 0.051

When the switch is flipped to B the current through the resistor and the inductor still remains same i = 8.7 amps nd it decays with the time const L/(R1 + R2)

current after 0.25 s

i(t) = 8.7exp(-t(R+R2)/L

       = 8.7exp(-2.5t)

di/dt = 8.7*-2.5exp(-2.5t) = -21.75exp(-2.5t)

voltage across inductor = Ldi/dt

                                  = -5*21.75exp(-2.5*0.25) = -58.21 V

i(0.25) = 8.7 exp(-0.25*12.5/5) = 4.66 mp