JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II PRINCIPLES OF ELECTRICAL ENGINEERING Previous Question Paper (November/December - 2020 - R15)

For a series RC circuit connected to a DC voltage source V at time t = 0, applying Kirchhoff's Voltage Law (KVL) yields:

Differentiating with respect to time t gives the linear differential equation:

Rearranging the terms leads to:

Integrating both sides results in the general solution for current:

At t = 0, the capacitor acts as a short circuit, so the initial current is I₀ = V/R. Substituting this into the equation, we find A = V/R. Thus, the expression for the transient current is:

Where RC is the time constant (τ) of the circuit. The current response waveform starts at a peak value of V/R at t = 0 and decays exponentially toward zero as the capacitor charges and reaches the steady state where the circuit acts as an open circuit. The waveform is a downward-sloping exponential curve starting at (0, V/R) and approaching the time axis asymptotically.

For an RL series circuit with a resistance R and inductance L connected to a DC voltage source V at t = 0, the circuit is governed by Kirchhoff's Voltage Law: V = Ri(t) + L(di(t)/dt). Applying the Laplace transform or solving the first-order linear differential equation with the initial condition i(0) = 0, the current response is i(t) = (V/R)(1 - e^(-Rt/L)).

The voltage across the inductor v_L(t) is given by v_L(t) = L(di(t)/dt). Substituting the expression for current:

The voltage across the resistor v_R(t) is given by v_R(t) = R * i(t):

At t = 0+, the inductor acts as an open circuit, so v_L(0+) = V and v_R(0+) = 0. As t approaches infinity, the inductor acts as a short circuit, so v_L(infinity) = 0 and v_R(infinity) = V.

The waveform for the voltage across the inductor v_L(t) starts at V and decays exponentially towards zero with a time constant τ = L/R. The waveform for the voltage across the resistor v_R(t) starts at 0 and rises exponentially towards V, following the same time constant τ.