Question

Review | Constants Part A After the switch in the circuit in (Figure 1) has been open for a long time, it is closed att0 Calculate the initial value of i Express your answer to three significant figures and include the appropriate units. View Available Hintls) Figure 1 of 1> imitialValue Units Submit 20 kΩ 200 mH 75 kΩ ▼ Part B 50 mA Calculate the final value of i Express your answer to three significant figures and include the appropriate units.

Answer 1

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For t < 0: - Assume steady state is reached before t = 0. The circuit before t = 0^- is Because At steady state inductor act as short circuit Applying source transformation technique Apply nodal analysis at node 'a' V - 250/25k + V/50k + V - 75/75k = 0. 11/150 V = 11 V = 150 V Therefore i_L = i (0^-) = V - 250/25k = 150 - 250/25k = 4 mA Part A: - i_initial = i (0^-) = 4 mA = 4.000 mA For t > 0: - The circuit is For t = infinity: The steady state is reached. The circuit is Apply source transformation technique. Because current prefer least resistance, the total current flows through short circuit path Therefore i (infinity) = i_final = 1 mA part B: - i_final = i (infinity) = 1 mA = 1.000 mA For t > 0: - time constant T = L_eq/R_eq L_eq = 200 mH For Find R_eq: - consider the circuit shown in figure R_eq = (50 k Ohm Vertical-bar Vertical-bar 75 k Ohm) = 30k Ohm Therefore tau = L_eq/R_eq = 200m/30k = 6.666 mu sec Part C: - tau = 6.666 mu sec