Question

In the circuit shown, the switch is opened at $t = 0$ s. The current $i(t)$ at $t = 2$ ms is ________ mA (rounded off to two decimal places).

 

Hint:

For transient analysis of circuits with inductors, determine the initial and final conditions of the current through the inductor and the time constant of the circuit after the switching event. The current will follow an exponential curve between the initial and final values.

Answer 1

Step 1: Determine the initial current through the inductor $i_L(0^-)$.
(Detailed calculation in previous attempts yielded $i_L(0^-) = 75/17$ mA)
Step 2: Determine the initial current through the $120 \Omega$ resistor $i(0^-)$.
(Detailed calculation in previous attempts yielded $i(0^-) = 25/17$ mA)
Due to continuity, $i_L(0^+) = i_L(0^-) = 75/17$ mA.
Step 3: Analyze the circuit for $t>0$ to find the final steady-state current through the $120 \Omega$ resistor $i(\infty)$.
(Detailed calculation in previous attempts yielded $i(\infty) = 5/6$ mA)
Step 4: Determine the Thevenin equivalent resistance seen by the inductor for $t>0$.
(Detailed calculation in previous attempts yielded $R_{eq} = 85.52 \Omega$)
Time constant $\tau = L/R_{eq} = 0.2 / 85.52 = 0.002338$ s.
Step 5: Write the expression for the current through the $120 \Omega$ resistor $i(t)$.
$i(t) = i(\infty) + (i(0^+) - i(\infty)) e^{-t/\tau}$
$i(t) = 0.833 + (1.47 - 0.833) e^{-t/0.002338}$
$i(t) = 0.833 + 0.637 e^{-t/0.002338}$
Step 6: Calculate $i(2 { ms})$.
$i(0.002) = 0.833 + 0.637 e^{-0.002/0.002338} = 0.833 + 0.637 e^{-0.855}$
$i(0.002) = 0.833 + 0.637 \times 0.425 = 0.833 + 0.2707 = 1.1037$ mA.
Step 7: Round off to two decimal places.
$i(2 { ms}) \approx 1.10$ mA.