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.
Show Hint
Inductor current cannot change instantly; use its pre-switch value as a fixed current fed into the new parallel resistor combination.
Step 1: Find the steady-state condition before switching (S open).
With S open, only the source, the inductor, and the first 100 ohm resistor form a closed loop; the second 100 ohm resistor is disconnected since the switch in series with it is open.
In DC steady state, an inductor carries constant current and behaves like a plain wire (its voltage drop is \(V_L = L\frac{di}{dt} = 0\) once the current stops changing). So all 10 V of the source appears across the 100 ohm resistor, and the inductor current is:
\[ I_L(0^-) = \frac{10\text{ V}}{100\ \Omega} = 0.1 \text{ A} \]
Step 2: Apply the inductor's continuity rule at the switching instant.
An inductor resists a sudden jump in current, because that would need an infinite voltage across it. So the current through the inductor cannot change in zero time:
\[ I_L(0^+) = I_L(0^-) = 0.1 \text{ A} \]
This 0.1 A must still be flowing out of the inductor into node \(V_A\) at the instant right after S closes.
Step 3: Work out the resistor network right after switching.
The moment S closes, the second 100 ohm resistor is connected in parallel with the first 100 ohm resistor, both running from node \(V_A\) to the ground rail. Their parallel combination is:
\[ R_{eq} = \frac{100\times 100}{100+100} = 50\ \Omega \]
Step 4: Compute \(V_A\) at \(t=0^+\).
The inductor is still forcing 0.1 A into node \(V_A\) (it cannot jump instantly to a new value), and this current now flows through the 50 ohm equivalent resistance to ground, so:
\[ V_A(0^+) = I_L(0^+)\times R_{eq} = 0.1\times 50 = 5 \text{ V} \]
Step 5: Final Answer.
Immediately after the switch closes, \(V_A = 5\) V.
\[ \boxed{5 \text{ V}} \]
