So I am having a hard time trying figuring this out. The question asks for the current through the 20Ω resistor as well as the current through the battery AFTER THE SWITCH HAS BEEN CLOSE FOR ALONG TIME.
Here is a picture of the circuit.
Kirchhoff Loop rule.
Kirchhoff node rule.
V_L = L*di/dt
i = I (e^(-t(R/L)))
The Attempt at a Solution
Here is how I tried figuring it out. After a long time di/dt =0 so the potential across the inductor is 0. If that is the case then by Kirchhoff Loop rule (outer loop) E - 10Ω*I - V_L = 0 or 30 - 10Ω*I - V_L = 0
since V_L = 0, then the current through the resistor (and the battery) would be 3A. Now, when I appply the loop rule on the right loop. V_L - 20Ω(I) = 0. Again, Since V_L = 0 then the current through the 20Ω resistor must be zero, but this doesn't make any sense to me, i just don't know why but I feel this is wrong. Can anyone confirm this? Also, if this is right, why is the current through that resistor 0 ? Thank you