1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Steady-state current

  1. Mar 31, 2010 #1
    i have an inductive-resistive circuit with Vbat = 6.0V, R = 10 ohm and L = 0.10mH

    I need to determine the steady-state current and the magnitudes of the steady-state voltages across the resister and across the i ductor.

    the switch has been closed at t=0s

    is the equation i need i(t) = (Vbat/R)(1-e^-Rt/L)

    and if t=0 do i just use 0 in place of t in the equation or do i use a time constant i worked out previously to be 0.01t?
  2. jcsd
  3. Mar 31, 2010 #2
    A few points here:
    what are you measuring the time constant in? you say 0.01t....what units are those? Units are very important when you want to give a physical meaning to your calculations.

    Secondly, the steady state current is just the theoretical maximum of the current as it builds up as far as it can go. It's a simple relationship between the maximum potential of the battery and the resistance of the circuit.
    In order to answer the second part about voltage across inductor and resistor, try thinking about what these two elements will be doing when you reach the steady state current...you don't even really need to calculate that part.

    Hope that helps
  4. Mar 31, 2010 #3
    I forgot to say: the equation you gave gives the current as a function of time. It's not really necessary if you're trying to find the steady state current, but you can use that equation if you like. Just take time (t) to the limit and see what happens to the right-hand side of the equation.
  5. Mar 31, 2010 #4
    thanks for quick reply, this is due tommorow :(

    I got the 0.001t because previously in the question it had asked me to work out the time constant in the circuit.

    I have found out how to work outthe time when the current reaches 50% of steady state value. I believe that to use the equation 0.5=e^-Rt/L

    I am really confused about this though
  6. Mar 31, 2010 #5
    ok...firstly, the time constant units should be in seconds...so if you calculated 0.01s you should check what your starting units are in. HINT: if you travel 3 kilometers in 1 second you can't call that 3 m/s....

    imax = Vbat/R is the equation you need to find the steady state current. At this point, the voltage will be fully across the resistor. That should be enough information to help you find the voltage across the resistor and the inductor.

    To find the time at which the current reaches 50% of its steady state value, first you need to find the maximum current (steady state current) using the formula I showed you above. Then, use the formula that you gave in your original post, which shows i as a function of time, and put them together. You know that half of the steady state value (on the left hand side) will be equal to the equation for i as a function of time at time t_half. Then just rearrange the equation to find t_half. Make sense?
  7. Mar 31, 2010 #6
    So the steady state will just be 6.0V/10ohm = 0.6V, the magnitudes across the resistor and the inductor i am not to sure about. Is it as simple as Vresistor = I/V
  8. Mar 31, 2010 #7
    is the current across the inductor i(t)=I(e(^-tR/L) ?
  9. Mar 31, 2010 #8
    Not quite. Your value is right, your units are wrong. You don't measure current in V.

    On the magnitudes of the voltage across the resistor and inductor...use Ohm's law: V=iR
    At steady state, the voltage will be all across the resistor.
  10. Mar 31, 2010 #9
    sorry its 0.6A
    So the magnitude over the resistor should be V=0.6A/10ohm=0.06V

    i think i have the wrong equation for the inductor
  11. Mar 31, 2010 #10
    "is the current across the inductor i(t)=I(e(^-tR/L) ?"

    The VOLTAGE across the inductor changes as a function of time. What are you looking for, current or voltage? There is a graph here that shows how voltage across the inductor changes as a function of time:
  12. Mar 31, 2010 #11
    I am looking for the voltage across the inductor. I cant seem to find in my text how to work this out.
  13. Mar 31, 2010 #12
    The point is that the voltage across the inductor and the resistor change as a function of time. You can't answer the question you posed unless you say at what point you're interested in.
    If you look at the graph of inductor voltage / time in that link I posted, you'll see what happens to the voltage across the inductor over time. Then ask yourself what steady state means. That should give you an answer without needing much calculation. You can also use that answer to deduce the voltage across the resistor.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Steady-state current