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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 $$V_{S1} = -10 V , V_{S2} = 20 V$$

Circuit Diagram

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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – Dave Tweed Oct 9 '12 at 20:50
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  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.

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  • \$\begingroup\$ 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 \$\endgroup\$ – 40Plot Oct 9 '12 at 20:53
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    \$\begingroup\$ 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. \$\endgroup\$ – The Photon Oct 9 '12 at 23:28

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