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).
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).
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Correct Answer: 4.6
Solution and Explanation
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.
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