Solving First Order RL circuit

Discussion in 'Homework Help' started by TripleDeuce, Mar 16, 2011.

  1. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010

    At time t = 0_

    The circuit looks like the voltage source in series with R1 and R2 with the inductor acting like a short circuit.

    iL (0_) = 60V/50Ω = 1.2A

    At t = 0+

    The circuit looks like

    Would iL(0 +) be a different current or the same? I have a feeling it would be the same.

    It is true that the inductance current remains the same right before and right after the switch is activated because the current through the inductor can't change instantaneously right?

    Any help on solving fr IL( 0+) and iL(t) would be appreciated
  2. syed_husain

    Active Member

    Aug 24, 2009
    yes, iL(0-)=iL(0+). now just apply node analysis. consider node "a" ground to make calculation easier. u will able to find the iL(t). hope this helps.
  3. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
    I make a node called V1 where the three resistors intersect each other and another node called VL above the inductor

    The node equation for V1 is

    (V1-60)/30 + V1/60 + (V1-VL)/20 = 0

    V1/30 - 2 + V1/60 + V1/20 - VL/20 = 0

    The node equation for VL is

    (VL-V1)/20 + 1.2 = 0

    VL/20 = V1/20 - 1.2

    VL = V1 -24

    Plugging that back in the node equation for V1

    V1/30 - 2 + V1/60 + V1/20 - (V1-24)/20 = 0

    V1/30 - 2 + V1/60 + V1/20 - V1/20 + 24/20 = 0

    V1 = -24

    so VL = -24 - 24 = -48V ?

    I have a feeling something is wrong.
  4. syed_husain

    Active Member

    Aug 24, 2009
  5. Vahe


    Mar 3, 2011
    Since this is a first-order transient problem, the solution for the inductor current for  t \ge 0will be of the following form

    <br />
i_L(t) = i_L(\infty) + [ i_L(0^+) - i_L(\infty) ] e^{-t/\tau}<br />

    where  i_L(0^+) is the current in the inductor right after the switch is closed. Since there are no impulse sources in the circuit, the inductor current is continuous and therefore i_L(0^+)= i_L(0^-). Current i_L(0^-) is the inductor current right before the switch is closed. The current i_L(\infty) is the value of the inductor current after the switch has been closed for a long time. The time constant \tau=L/R_{th}, where R_{th} is the Thevenin resistance seen by the inductor in the circuit after the switch is closed.

    Looking at the circuit before the switch is closed and assuming that the switch has been open for a long time, we can assume dc steady steady (L is a short circuit); therefore,

    <br />
i_L(0^-) = \frac{V_s}{R_1+R_2}=\frac{60V}{50\Omega}=\frac{6}{5}\text{A}<br />

    Since the inductor current is continuous over the switching action. We have that i_L(0^+) = \frac{6}{5}\text{A}. After the switch is closed for a long time, we reach another dc steady state and the inductor is once again a short circuit. First we determine the current i_s out of the + terminal of the voltage source as

    <br />
i_s = \frac{V_s}{R_1 + R_2 || R_3}=\frac{60V}{30\Omega+15\Omega}=\frac{4}{3} \text{A}<br />

    This is also the current going through R_1 from left to right. Now with the inductor being a short circuit, the current that we seek can be given by current division.

    <br />
i_L(\infty)= \frac{R_3}{R_2 + R_3} i_s=\frac{60\Omega}{80\Omega} \, (\frac{4}{3}) \text{A}=1\text{A}<br />

    The resistance R_{th} is found by turning off the voltage source and finding the resistance "seen" by the inductor.

    <br />
R_{th}=R_2 + R_1||R_3 = 20 + 30||60 = 40 \Omega<br />
\tau = L/R_{th} = 0.4mH/40\Omega = 10\mu s<br />

    Therefore the solution is

    <br />
i_L(t) = 1 + 0.2 e^{-100,000t} \text{  for } t \ge 0<br />
i_L(t) = 1.2 \text{  for } t < 0<br />

    I will leave you to calculate v_L=L di_L/dt :)
    Hopefully, I have not made any calculation errors :(