Question

Refer to the transient L-R circuit below. Answer / calculate the following, showing ALL steps R1 65Ω S1 R4 600Ω 20V 1.2k2 1.2k R3 L1 50Ω 30uH a) Deduce the mathematical expression for the current through L1 if the charging process starts at t: 0.005 us (Approx. 0.00 μs) b) Assume switch S1 is switched over at t 0.3 us to stop the charging and start the discharging process. Calculate the magnitude of the current through the inductor L1 at the instant the switch is opened curve where it stops as calculated in (b) expression for the current through L1 if the discharging process starts at c) Draw a graph of the possible charging curve from (a) and the actual d) Now determine the discharging time constant and the mathematical t- 0.45 H e) Draw a graph of the charge / discharge process (inductor current vs. f) Obtain the expression for the inductor charge voltage vs. time by g) Now obtain the expression for the inductor discharge voltage vs. time by time differentiating the expression for charge current and multiplying by L1 differentiating the expression for discharge current and multiplying by

Answer 1

Given transient L-R circuit as shown below. Now, we deduce the mathematical expression the current through L_1. Let us consider the circuit from the L-R circuit. Apply KVl to the above circuit. 20 = 50 I_1 + 65 I_1 + L dI_1/dt. \r\n Because L dI_1/dt = voltage across Inductor. 20 = 115 I_1 + 30 mu dI_1/dt 30 times 10^-6 dI_1/dt = 20 - 115 I_1 integral dI_1/20 - 115 I_1 = 30 tiems 10^5 integral dt. Ln (20 - 115 I_1)/-115 Vertical-bar ^I_1_0 = 30 times 10^5 t Vertical-bar ^t_0.005 [Because here initial condition is 0.005] 20 - 115 I_1/-115 = e^-30 times 10^-6 t - e^-(30 times 10^-6) times 0.005 I_1 = 115 - 115 e^-30 times 10^-6 t - 115/115 A Now, We calculate the magnitude of the current through the inductor L_1 at the instant the switch is opened At t = 0.3 mu s. I_1 = -115 e^-30 times 10^-6 t/115 = -115 e^-30 times 10^-6 (0.3 mu)/115 I_1 = 1 A