Consider an L-R circuit as shown in the figure. (Figure 1) The battery provides
ID: 1646864 • Letter: C
Question
Consider an L-R circuit as shown in the figure. (Figure 1) The battery provides 12.0 V of voltage. The inductor has inductance L, and the resistor has resistance R = 150 Ohm. The switch is initially open as shown. At time t = 0, the switch is closed. At time t after t = 0 the current I(t) flows through the circuit as indicated in the figure. The current in the circuit will approach a constant value I_c after a long time (as t tends to infinity). What is I_c? To find the value of the current after a long time, you need to evaluate the exponential expression for I(t) at a large value of t. Mathematically, that corresponds to evaluating the exponential function at infinity. When you do that, you will get e^-infinity, which is equal to zero.Explanation / Answer
applying KVL in loop
V - LdI(t)/dt - I(t)*R = 0
=> dI(t)/dt + (R/L)I(t) = V/L
this is differential equation
=> I(t) = V/R*( 1 - e^(-Rt/L) )
at t = very long time = infinite
=> I(t) = V/R = 12/150 = 0.08 A = Ic
OR IN SHORT AFTER VERY LONG TIME INDUCTOR ACTS AS A SHORT CIRCUIT .
=> THERE IS ONLY RESISTANCE INTHE CIRCUIT.