A 10H inductor has a resistance of 5Ohms. The inductor is connected in series with an open switch. The supply is 100V dc. A 'bleed' resistor of 20 Ohms is connected in parallel. CalculateA) The time taken for the inductor current to reach 2A.(Assume initial switch on.)B) With the supply removed, and the inductor current 'steady', the current value after 0.3s.The bleed resistor has completely thrown me off with this questionI asked this earlier, but the answers given didn't explain to me well enough, and i like to understand the answer.
I answered this before. If you are an EE student, you need to memorize the response of first order RL and RC circuits for step and impulse excitations and with initial conditions. You should have also studied and solved the differential equations that describe these circuits. Specifically, what don't you understand? When the switch is closed, the supply is in parallel with the 20 ohm resistor and (in parallel) with the inductor and its series resistor. I gave the expression for the inductor current. When the switch open the supply is removed and the 20 ohm resistor is in series with the inductor and its series resistance. I also gave you the expression.
Bleeding resistor won't affect current rise time Charging operation Timeconstant ? L/R 10/5 2 seconds Final current V/R 100/5 20A 2A 2/20 0.1 in Pu 0.1 1-e^-K e^-k 0.9 -k*ln e ln 0.9 -k*1 -0.105 k 0.105 t/? t 0.105? 0.21 sec discharging operation Timeconstant ? L/(R+20) 10/25 0.4 seconds Final current V/R 100/25 4 A Ipu 1*e^-0.3 2.7182^-0.3 0,741 I 0,741*4 2.963 A
