JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year BASIC ELECTRICAL ENGINEERING Previous Question Paper (November/December - 2020 - R18)

For an RL circuit with a resistor R and inductor L in series connected to a DC voltage source V at time t = 0, the governing differential equation is:

This is a first-order linear differential equation. The solution consists of a complementary function (transient response) and a particular integral (steady-state response). The general solution for the current i(t) is:

Key components of this response include:

• Steady-state current: As t approaches infinity, i(t) reaches V/R.
• Time constant (τ): Defined as L/R, it represents the time taken for the current to reach approximately 63.2% of its final steady-state value.
• Transient response: The term -(V/R)e^(-Rt/L) represents the decaying transient that accounts for the initial energy storage characteristics of the inductor.

To calculate the current in the 2 Ω resistor, apply nodal analysis or mesh analysis based on the provided circuit parameters. Assuming a standard circuit configuration where the 2 Ω resistor is part of a network with voltage sources and other resistors:

1. Define the nodes and assign node voltages, setting one node as the reference (ground).
2. Apply Kirchhoff's Current Law (KCL) at each principal node. For a node connected to the 2 Ω resistor, the sum of currents leaving the node must equal zero.
3. Solve the resulting system of linear equations to find the voltage across the 2 Ω resistor.
4. Calculate the current using Ohm's Law:
   I = V_resistor / 2

If the circuit is reducible, apply Thevenin's theorem by removing the 2 Ω resistor, finding the open-circuit voltage (Vth) and the equivalent resistance (Rth) seen from the terminals, then calculating the current as I = Vth / (Rth + 2).