A

Step 1: Consider a simple RC circuit with a resistor (R) and capacitor (C) in series with an ideal voltage source (E). When the switch is closed, the capacitor initially behaves like a short circuit because it has not yet charged.

Step 2: According to Ohm's Law \( V = IR \), the total current in the circuit is given by \( I = \frac{E}{R} \). At t=0 (the moment the switch is closed), the capacitor has no charge and acts as a short circuit, allowing maximum current to flow through the circuit.

Step 3: Therefore, the initial current, \( I_0 \), is equal to \( \frac{E}{R} \) immediately after the switch is closed.

Step 4: As time progresses, the capacitor begins to charge, and the current will decrease exponentially to zero as the capacitor becomes fully charged (steady state). However, the time it takes to reach this state depends on the time constant \( \tau = RC \).

Conclusion: Hence, it can be concluded that the current in the circuit does rise instantly to its steady state value \( \frac{E}{R} \) when the switch is closed.