The circuit shown in the following figure is initially under steady-state condition. The switch is moved from position 1 to position 2 at t = 0. The current after switching will be:

9) The circuit shown in the following figure is initially under steady-state condition. The switch is moved from position 1 to position 2 at t = 0. The current after switching will be:

1. 2e^-5t A [Option ID - 2865]
2. 1 - 2e^-5t A [Option ID - 2866]
3. 1 + 2e^-5t A [Option ID - 2867]
4. 2e^5t A [Option ID - 2868]

Answer: A

Solution

Step 1: Initial State (t < 0)

Before the switch moves, it is at position 1. In a DC steady-state condition, an inductor acts as a short circuit. The current flows only through the voltage source (20V) and resistor R₁ (10Ω).

i(0^-) = V / R₁ = 20V / 10Ω = 2 A

Step 2: The Switching Property

A fundamental rule of inductors is that the current cannot change instantaneously. Therefore, the current immediately after the switch moves (t = 0⁺) must be the same as it was before.

i(0⁺) = i(0^-) = 2 A

Step 3: Analyzing the Circuit for t > 0

At position 2, the voltage source is removed. We now have a "source-free" RL circuit where the inductor discharges its stored energy through both resistors R₁ and R₂ in series.

• Total Resistance (R_total) = R₁ + R₂ = 10Ω + 10Ω = 20Ω

Step 4: Applying the Discharge Formula

The standard formula for current decay in an RL circuit is:

i(t) = i(0) · e^-(R/L)t

Given the options provide an exponent of -5t, we can see that the ratio R/L must be 5. Substituting our initial current of 2A into the formula:

i(t) = 2e^-5t A

Final Answer:
The current after switching is 2e^-5t A. This corresponds to Option 1 [Option ID - 2865].

Scenario: Replacing Inductor with a Capacitor

If the inductor (L) were replaced by a capacitor (C), the circuit logic shifts from managing current to managing voltage.

1. Initial State (t < 0)

In DC steady state, a capacitor acts as an open circuit (no current flows). Since no current flows through R₁, there is no voltage drop across it.

V_C(0^-) = Source Voltage = 20V

2. At the moment of switching (t = 0)

The voltage across a capacitor cannot change instantaneously. When the switch moves to position 2, the capacitor initially behaves like a 20V battery.

V_C(0⁺) = V_C(0^-) = 20V

3. Post-Switching Current (t > 0)

In position 2, the capacitor discharges through both resistors (R₁ + R₂ = 20Ω). The initial current is found using Ohm's Law:

i(0⁺) = V_C / (R₁ + R₂) = 20V / 20Ω = 1 A

Comparison: Inductor vs. Capacitor

Feature	Original (Inductor)	Modified (Capacitor)
Steady State (t < 0)	Short Circuit (I = 2A)	Open Circuit (V = 20V)
What is preserved?	Current (2A)	Voltage (20V)
Initial Current (t = 0+)	2 A	1 A
Time Constant (τ)	L / R	R × C
Typical Equation	i(t) = 2e-(R/L)t	i(t) = 1e-t/(RC)

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