Question

A 9.5-V battery is connected in series with a 44-mH inductor, a 180-Ω resistor, and an open switch.

1. What is the current in the circuit 0.120 msms after the switch is closed? Express your answer using two significant figures.

2. How much energy is stored in the inductor at this time? Express your answer using two significant figures.

Answer 1

R L no E-9.5V L = 44 mH: 44X10-3H R 1800 t = 0.120 m/sec ε At time t, current is given as any By KVL IR +LDI ICT) = & (1 - 6tR/L) -E R I dI E-IR •dt Put all values and we get 0:/20x10-3x180 44x10-3 180) - Itt) = 79.5 (268) (1- l-e ln (8-IR) = - 4 Re-arrange IC) = (i-e tr/4) it I(t) = 0.02047 A I (t) = 20.47 MA (6) Stored Energy in the inductor is Us 1 LI2 at t= 0.120msee. U = {(448103H) ( 20.47X10 3A)2 U = 9. 22 *1867 U= 9.22 MJ