Question 9.18: The switch in Figure 9–11 has been in position A for a long ......
The switch in Figure 9–11 has been in position A for a long time. At t = 0 it is moved to position B. Find iL(t) for t ≥ 0.
S T E P 1 The circuit differential equation is found by combining the KVL equation and element equations with the switch in position B.
KVL: υ_{R}(t) + υ_{L}(t) = 0
Resistor: υ_{R}(t) = i_{L}(t)R
Inductor: υ_{L}(t) = L\frac{di_{L}(t)}{dt}
Substituting the element equations into the KVL equation yields
L\frac{di_{L}(t)}{dt} + Ri_{L}(t) = 0
Prior to t = 0, the circuit was in a dc steady-state condition with the switch in position A. Under dc conditions the inductor acts like a short circuit, and the inductor current just prior to moving the switch is iL(0−) = I0 = VA/R.
S T E P 2 Using the linearity and differentiation properties, we transform the circuit differential equation into the s domain as
L[sIL(s) – I0 ] + RIL(s) = 0
Solving algebraically for IL(s) yields
I_{L}(s) = \frac{I_{0}}{s + R/L} A-s
S T E P 3 The inverse transform of IL(s) is an exponential waveform:
i_{L}(t) = \left[I_{0}e^{−Rt/L}\right] u(t) A
where I0 = VA/R. Substituting iL(t) back into the differential equation yields
L\frac{di_{L}(t)}{dt} + Ri_{L}(t) = −RI_{0}e^{−Rt/L} + RI_{0}e^{−Rt/L} = 0
The waveform found using Laplace transforms does indeed satisfy the circuit differential equation and the initial condition.