# First Order RL Circuit

In figure 8.21 $$R_1 = 50 , R_2 = 200 , L = 2 H$$ Voltage is given by $$V_{in}(t) = V_{s1}u(-t) + V_{s2}u(t)$$ where $$V_{S1} = -10 V , V_{S2} = 20 V$$ a) Find $$I_L(0^+)$$ and $$I_L(t)$$ for t > 0

So far, I haven't done anything because I don't know what to do first. Some hints would appreciated.

• What's the significance of the switch "S"? Are we supposed to assume that it opens at t=0? If so, you can ignore R2 altogether, since it never has any current going through it. If not, start by replacing the voltage source and the two resistors with their Thevenin equivalent. – Dave Tweed Oct 9 '12 at 20:50

## 1 Answer

1. Knowing that an inductor will initially behave as an [open][short](pick one) circuit will answer the behavior at t=0+.

2. Knowing how the inductor behaves at t = infinity will let you find the asymptotic value of IL.

3. Figuring out the time constant of the circuit will let you fill in the time in between.

• At t=0 the inductor will behave as an open circuit. and just wondering , what about at t=0- (t 0*minus) how does it differ from t 0*plus – 40Plot Oct 9 '12 at 20:53
• The main thing you need to know is, the current through an inductor will never change instantly. Similarly, but not relevant to this problem, the voltage across a capacitor will never change instantly. – The Photon Oct 9 '12 at 23:28