Question

The switch $S_1$ was closed and $S_2$ was open for a long time. At $t=0$, switch $S_1$ is opened and $S_2$ is closed, simultaneously. The value of $i_c(0^+)$, in amperes, is _____________

• A. 1
• B. $-1$
• C. 0.2
• D. 0.8

Hint:

At switching instants: $v_C$ is continuous, $i_L$ is continuous. Use those to compute instantaneous branch currents, then apply KCL with the given current direction convention.

Answer 1

The Correct Option is B

Step 1: Steady state just before switching ($t=0^-$).
$S_1$ closed, $S_2$ open $\Rightarrow$ the inductor branch is disconnected (so $i_L(0^-)=0$).
DC steady state: capacitor is open; the 1-A source feeds the parallel $100\ \Omega$ and $25\ \Omega$ resistors.
$R_{\text{eq}} = (100 \parallel 25) = 20\ \Omega \Rightarrow V_{\text{node}}(0^-) = 1 \times 20 = 20\ \text{V}$.
Hence, $v_C(0^-) = 20\ \text{V}$.

Step 2: Initial conditions at $t=0^+$.
Capacitor voltage cannot change instantly $\Rightarrow v_C(0^+) = 20\ \text{V}$.
Inductor current cannot change instantly; since the branch was open, $i_L(0^+) = i_L(0^-) = 0$.

Step 3: Currents in the resistors at $t=0^+$.
With $S_1$ open and $S_2$ closed, the node sees $100\ \Omega$, $25\ \Omega$, and the inductor to ground.
At the instant $t = 0^+$, $v_{\text{node}} = 20\ \text{V}$, so
$I_{100} = \frac{20}{100} = 0.2\ \text{A},\qquad I_{25} = \frac{20}{25} = 0.8\ \text{A}$.
$I_L(0^+) = 0.$

Step 4: KCL at the node to find $i_c(0^+)$.
Take currents from node to ground as positive; capacitor current $i_c$ is defined downward (to ground).
With no external source connected, the capacitor must supply the resistive/inductive branch currents, so
\[ i_c(0^+) = -\big(I_{100} + I_{25} + I_L(0^+)\big) = -(0.2 + 0.8 + 0) = -1\ \text{A}. \]

Final Answer:
\[ \boxed{i_c(0^+) = -1\ \text{A}} \]