In the circuit shown in the figure, the battery is ideal, e = 13.9 V, L = 11.5 m
ID: 1453699 • Letter: I
Question
In the circuit shown in the figure, the battery is ideal, e = 13.9 V, L = 11.5 mH, R_1 = 12.2 and R_2 = 20.7 Ohm. The switch has been open for a long time before it is closed at t = 0 At what rate is the current in the inductor changing (a) immediately after the switch is closed and (b) when the current in the battery is 0.530 A. (c) What is the current (in A) in the battery when the circuit reaches its steady-state condition? By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor.Explanation / Answer
a) immediately after switch is closed
=> voltage across inductor = 13.9 * (12.2/32.9)
= 5.154 V
=> V = - L * di/dt
=> 5.154 = 11.5 * 10-3 * di/dt
=> di/dt = 448.17 A/sec ------------------> rate of change of current in inductor
b) when current in battery is 0.530 A
=> voltage across inductor = 2.929 V
=> 2.929 = 11.5 * 10-3 * di/dt
=> di/dt = 254.69 A/sec ------------------> rate of change of current in inductor
c) in steady state condition
current in battery = 13.9/20.7 = 0.671 A