Hello, i really need some help solving. please show all steps. i really appreciate it! thanks in advance.An 8.0-mH inductor and a 2.0-Ω resistor are wired in series to a 20-V ideal battery. A switch in the circuit is closed at time 0, at which time the current is zero. After a long time the current in the resistor and current in the inductor are: A) 0,0B) 10A, 10AC) 2.5A, 2.5AD) 10A, 2.5AE) 10A, 0
Assuming the inductor is a perfect conductor, the as long as the time is very long (t L/R), the current is set by the value of the resistor: I V/r 20/2 10 A. Since this is a series circuit, the current in the resistor and inductor is the same value.
Don't know what H is but the circuit should be ten amps if the inductor has no resistance. That narrows it down to two answers.
for an ideal inductor the impedance for D.C.(supplied by battery) is zero as (angular frequency omega is zero). now when such a combination is connected to D.C transient current flows into the circuit. and the value of current at any instant is given by. IV/R(1-e^(-Rt/L)) as t is long the current is given by, lim t--infinity V/R(1-e^(-Rt/L)) V/R(1-(1/e)^infinity) as 1/e 0, (1/e)^infinity tends to zero. so IV/R(1-0) IV/R20/210 Amperes note that transient current either growing or decaying lies for a very short while. so if long time is given. you can directly put impedance of ideal inductor as zero and of ideal capacitor as infinite. hope this helps. so answer is B) 10A, 10A
B is the Answer because: At t 0, the inductor is modeled as an open circuit because an inductor provides very high impedance (reactance) for rapidly changing current. The majority of the voltage in the ckt right after the switch is closed appears across the inductor. We can write the equation for the current in the ckt as I(t) [20/2][1-e^-(t*R/L)] Notice that at t0, I(0) 10*[1- e^(0)] 10*[1-1] 0 For t very large the term e^-(t*R/L) approaches zero That leaves I(t) 10A. The current in the inductor and resistor are the same because they are in series.
