Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: RLC Black Box

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A two-terminal “black box” is known to contain an inductor L, a
    capacitor C, and a resistor R. On connecting a 1.5 V battery, 1.5 mA
    flows. When an AC voltage of 1V RMS at 60 Hz is connected, 10 mA
    RMS flows. As the frequency increases at a fixed 1 V RMS, the
    current reaches a maximum of over 100 A at 1 kHz. Sketch the circuit
    in the black box and find values for L, C and R.

    2. Relevant equations

    3. The attempt at a solution

    I can't figure out how to start this one because of the phrase "on connecting the battery". I think that must mean immediately after connecting the battery, but I don't know how the circuit could be set up if that was the case because no current can flow through the capacitor when there is zero frequency. Can someone help me see how the circuit is supposed to be set up?
  2. jcsd
  3. Jan 26, 2010 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    What would happen if the capacitor was connected in parallel to the resistor and/or inductor?

  4. Jan 26, 2010 #3
    I figured this: it can't be connected in parallel to the resistor or inductor + resistor because right after the battery is connected the potential drop across the capacitor is 0 because it hasn't charged yet, and the problem states there is current. Since there is current after the battery is connected there must be no potential across the inductor too right? Because the inductor won't allow sudden changes in current like going from 0 to 1.5 mA
  5. Jan 26, 2010 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    A 0 voltage across the capacitor does not mean there is no current in the circuit. Current leads voltage in a capacitor. So there is current through the capacitor initially as charge (and voltage) is building up. When there is maximum voltage drop across the capacitor there is no current at all.

    You have to assume that the current of 1.5 mA is the stable current that results from the application of 1.5 DC. The current becomes stable after a very short time so ignore initial effects.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook