The circuit given below is initially in steady-state with the switch S open. Immediately after the switch S is closed, the voltage \(V_A\) is ________ volts.
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Inductor current cannot change instantly; use its pre-switch value as a fixed current fed into the new parallel resistor combination.
Step 1: Model the inductor as a current source at the switching instant.
A key property of an inductor is that the current through it is a state variable, it cannot jump. So at the exact instant the switch changes, we can freeze the inductor's current at whatever value it held just before, and treat the inductor as if it were replaced by an ideal current source of that fixed value, feeding current into node $V_A$.
Step 2: Find that frozen current from the pre-switch steady state.
Before S closes, the circuit is just the 10 V source, the inductor (a short circuit in steady state), and the first 100 ohm resistor in one loop. So the inductor's steady current is:
\[ I_0 = \frac{10}{100} = 0.1 \text{ A} \]
This is the value our equivalent current source will carry at $t = 0^+$.
Step 3: Redraw the circuit at $t=0^+$ using this current source.
At $t = 0^+$, node $V_A$ is fed by a 0.1 A current source (standing in for the inductor), and this node connects to ground through the first 100 ohm resistor AND, now that S is closed, also through the second 100 ohm resistor. These two resistors sit side by side between $V_A$ and ground, so from the current source's point of view they act as one parallel resistor:
\[ R_{parallel} = 100 \parallel 100 = 50\ \Omega \]
Step 4: Apply Ohm's law at the node.
A current source forces its full current through whatever resistance it faces, so the node voltage is simply that current times the resistance it sees:
\[ V_A = I_0 \times R_{parallel} = 0.1 \times 50 = 5 \text{ V} \]
Step 5: Final Answer.
Treating the inductor as a frozen 0.1 A current source immediately after switching gives $V_A = 5$ V, the same result as the direct node analysis.
\[ \boxed{5 \text{ V}} \]
