On part a of this problem, is i(0+) = 0 since the circuit is open and no voltage is flowing? Then i(inf) = 0 also? Or no? Could someone please help me?
Hi, As you must be knowing,the inductor opposes any change in the current flowing through it,so the current at t=0+ is same as the current flowing at t=0-,which is 8V/8Ω=1A. At t=∞,the current flowing is 0
I used this equation to obtain τ = L/R = 8/8 = 1 I have: i(0) = 1A i(∞) = 0 So for i(t) from the equation above, I get: i(t) = 0 + (1 - 0)e^(-t/1) = e^(-t) Is that correct?
Hi, Solution to part a seems to be correct. In part b,observe that the switch is first open for a long time,as a result the current flowing through the circuit is zero.Now remember as I told you,the inductor opposes any change in the current flowing through it,so as soon as you close the switch the inductor will not allow any current to flow through it.So current through inductor at t(0+)=0.All the current at that instant will flow through the two 8Ω resistors.The total resistance will be 8+8.The current will be 0.5A. At time t=inf,the circuit will reach the steady state,inductor will behave as short circuit,so the only resistance coming into picture will be 8 ohm,resulting in current of 1A
Inductors oppose an instantaneous change in current. So in part b) your inductor will appear as an open circuit when t = 0+. They achieve this by either a collapse or growth of their magnetic field, which in the case of part b) causes a potential difference across the inductor that is equal and opposite to the source voltage, hence no current flow. In part a), the inductor would have a potential difference of -8V across it at t=0+, making it act like a source of 8V, this is clearly because an 8V voltage source has been removed, so to make the current not change instantaneously, the inductor tries to act like the source and give the current the same amount of energy it had at t=0-. Hope that makes sense. Capacitors oppose an instantaneous change in voltage, for your information.