2.In the figure, after the switch is closed at time t0, the emf of the source is automatically adjusted to maintain a constant current I through S. a) Find the current through the inductor as function of time. b) At what time the current through the resistor equal to the current through the inductorthe circuit looks like this:a constant current source with a switch in series and then splits into a resistor and inductor in parallel
After the switch is closed, the initial current will only flow through the resistor, since current through an inductor cannot change instantaneously. The steady state current will only flow through the inductor due to lower impedance. To make things easier, we can change the parallel current source and resistor to their thevenin equivalent series voltage source and resistor. Vth I*R Rth R So now the series equations will apply: v.L(t) (IR)exp(-tR/L) L*di/dt i.L(t) I*(1-exp(-tR/L)) Resistor inductor currents are equal when: i.L(t) I*(1-exp(-tR/L)) 0.5*I Solving for time t gives: t -(L/R)ln(0.5)
the means in an inductor is proportional to the oblong of the present. At area the present it holds a million/4 the quantity of means. It held 40 J on the commencing, it now holds 10 J so the distinction is 30 J
