A 10 H inductor has a resistance of 5ohms. The inductor is connected in series with an open switch. The supply is 10v dc. A 'bleed' resistor of 20 ohms is connected in parallel with the coil. What is the time taken for the inductor current to reach 2assuming initial switch on
I suspect that the 20 ohm parallel resistor confused the picture for you. However, the 20 ohm, parallel resistor is independent of the current through the inductor, and therefore is not considered in this calculation. The current flow through the inductor is modeled as an independent, RL circuit with 5 ohm resistor in series with the 10 H inductor. The reasoning is that the current through this RL branch is only dependent on voltage across the series resistance and inductor. In the ideal parallel circuit, the voltage across the RL branch is source voltage at time t 0+ (instant of switch closure). Analyze the time dependent current as you would a simple RL circuit with the 5 ohm resistor in series and ignore the 20 ohm resistor. Analysis: V(t) V(r) + V(L) V(r) i * R V(L) L(di / dt) V(t) i * R + L(di / dt) i(t) (V/R)(1 - e^(-Rt / L)) then solve for 't' with i 2 amperes.
