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Find Va(0-) and Va(0+) with 2 H inductor

Short answer

Va(0⁻) = 3.33 V, Va(0⁺) = 40/21 ≈ 1.905 V.

In the network shown in the figure below, a steady state is reached with the switch open. At t = 0, the switch is closed. Determine the value of Va(0⁻) and Va(0⁺). [Figure: 5 V source. A 10 Ω resistor runs along the top wire from the source positive terminal to the right top node (bridging). A second row has a 10 Ω resistor from the source positive terminal to node Va, then a 20 Ω resistor from Va to the right node. A 2 H inductor connects the right node to the bottom wire. A switch (closes at t = 0) in series with a 10 Ω resistor connects Va to the bottom wire.]

Assumptions from the figure: 5 V source. A 10 Ω resistor connects the source + terminal directly to the right node R (the node at the top of the inductor). A second 10 Ω goes from the source + to node Va, then a 20 Ω from Va to R. The 2 H inductor connects R to ground. A switch in series with 10 Ω connects Va to ground and closes at t = 0.

At t = 0⁻ (switch open, dc steady state)

The inductor is a short, so V_R = 0. The path source → 10 Ω → Va → 20 Ω → ground carries

I = 5/(10 + 20) = 1/6 A
Va(0⁻) = 20 × (1/6) = 3.333 V

Inductor current: through the direct 10 Ω: 5/10 = 0.5 A, plus 1/6 A, so

i_L(0⁻) = 0.5 + 0.1667 = 0.6667 A

At t = 0⁺ (switch closed)

The inductor still carries 2/3 A (it acts as a 0.667 A current source into ground). Node equations, with Va and V_R unknown:

Node Va:  (5 − Va)/10 = Va/10 + (Va − V_R)/20
          2(5 − Va) = 2Va + (Va − V_R)  →  V_R = 5Va − 10
Node R:   (5 − V_R)/10 + (Va − V_R)/20 = 2/3
          2(5 − V_R) + (Va − V_R) = 13.333
          Va − 3V_R = 3.333

Substitute V_R = 5Va − 10: Va − 15Va + 30 = 3.333, so

−14 Va = −26.667
Va(0⁺) = 1.905 V (= 40/21),   V_R(0⁺) = −0.476 V

Answer: Va(0⁻) = 3.333 V and Va(0⁺) = 1.905 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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