In the circuit given below, V=26 V and R=1 ohm. The switch in the circuit has been closed for a long time but is opened at t =0. Determine i(t) for t>0.
The Correct Answer and Explanation is:
The correct answer is i(t) = 52e^(-1.5t) – 26e^(-3t) A.
Explanation
To determine the inductor current i(t) for t>0, we follow a three step process: analyze the circuit for t<0 to find initial conditions, characterize the circuit for t>0, and use the initial conditions to find the final solution.
First, for t<0, the switch has been closed for a long time, meaning the circuit is in a DC steady state. In this state, the inductor acts as a short circuit (a wire), and the capacitor acts as an open circuit. The inductor shorts out the parallel 2R resistor, so all the current passing through the first resistor R flows through the inductor. The initial inductor current is i(0) = V/R = 26V / 1 ohm = 26 A. The capacitor is in parallel with the inductor (which is a short circuit), so its initial voltage is v_c(0) = 0 V. Due to their physical properties, inductor current and capacitor voltage cannot change instantaneously, so i(0+) = 26 A and v_c(0+) = 0 V.
Second, for t>0, the switch opens, disconnecting the voltage source and the first resistor. The remaining circuit is a source free parallel RLC circuit consisting of the 2R resistor, the inductor L, and the capacitor C. The response of this circuit is described by a second order differential equation. We must determine the type of response by calculating the neper frequency (alpha) and the resonant frequency (omega_0). For a parallel RLC circuit, alpha = 1 / (2 * (2R) * C) = 1 / (4RC) = 1 / (4 * 1 * 1/9) = 2.25. The resonant frequency is omega_0 = 1 / sqrt(LC) = 1 / sqrt(2 * 1/9) ≈ 2.12. Since alpha > omega_0, the circuit is overdamped.
Third, the general solution for an overdamped response is i(t) = A1e^(s1t) + A2e^(s2t). The roots are s1,2 = -alpha ± sqrt(alpha^2 – omega_0^2), which calculates to s1 = -1.5 and s2 = -3. We use the initial conditions to find the constants A1 and A2.
From i(0) = 26 A, we get A1 + A2 = 26.
The second condition comes from di(0)/dt = v_c(0) / L = 0 V / 2 H = 0. The derivative is di/dt = -1.5A1e^(-1.5t) – 3A2e^(-3t), so at t=0, we have -1.5A1 – 3A2 = 0. Solving these two simultaneous equations gives A1 = 52 and A2 = -26. Substituting these constants yields the final expression for the current.
