Recent content by Fat_Squirrel

  1. Fat_Squirrel

    Charging a capacitor

    Cheers, too late now, missed my deadline. I can't believe all I needed was the product rule (what I'm currently doing in maths .. oh man). I'll play with this later in the week when I have a little free time. Thanks for the help.
  2. Fat_Squirrel

    Charging a capacitor

    One last request for help. I have an hour and a half until I need to submit this.
  3. Fat_Squirrel

    The formulas for parallel and series connections

    Here's my thinking, the two x 4s are essentially a single 2. So from that side it can go along either 2 ohm path. Then they'd cancel each other out going through the 5, so no current flows through that resistor.
  4. Fat_Squirrel

    The formulas for parallel and series connections

    I'd be thinking something like this : And keep simplifying parallel circuits til things look simpler. Again, warning, not my specialty, but something to think about. You seem to have an answer, see if you can keep reducing it in that fashion and see what you get.
  5. Fat_Squirrel

    The formulas for parallel and series connections

    Most of the current will take the easiest path, but that doesn't mean it all does. Eg, Circuit has branch, one has 2ohm resistor, the other 4. Do you really think it ALL goes via the easier path? Picture a supermarket queue. There's a fast moving 10 items or less queue, or a slow moving...
  6. Fat_Squirrel

    The formulas for parallel and series connections

    Looking at the original diagram, current can go through the first 4's then come back through the 5, This isn't possible on your diagram. (also it could then go through the 20 from the 5 - again, not possible on your diagram). Not sure what the question is asking but I'd start simplifying...
  7. Fat_Squirrel

    Charging a capacitor

    Is (b) requiring me to use ω CV Sin(ωt + pi/2) ??? Aside from that, I'm out of ideas. Anyone ??
  8. Fat_Squirrel

    Charging a capacitor

    Cheers, you're right, its 9.55 x 10^-7, so 0.955 uA. Any hints for b and d ??
  9. Fat_Squirrel

    Charging a capacitor

    For (C) I'm assuming its P=I^2 R = 9.55x10^-6 * 3000000 = 2.74 x 10^-6 Watts ??
  10. Fat_Squirrel

    Charging a capacitor

    Nothing I can find. Can't even google myself a formula. Closest guess would be - But I'm not even sure what dq are in this?
  11. Fat_Squirrel

    Charging a capacitor

    Is (b) just 1/2 QV, so 1/2 * 1/4 = 2 somethings ? And (d) just QV, so 1x4 = 4.
  12. Fat_Squirrel

    Charging a capacitor

    Just realised it only wants the rate, and not the rate of change of these things. So (a) we are looking for current, so E/R e^(-t/RC) = 4/3000000 * e^(-1/3) = 9.55uA ?? (b) energy I guess I use U = 1/2 CV^2 ? (c) Relates to Power, so P=I^2*R (d) can wait.
  13. Fat_Squirrel

    Charging a capacitor

    Homework Statement A 3.00 MΩ resistor and a 1.00 μF capacitor are connected in series with an ideal battery of emf E = 4.00 V. At 1.00 sec after the connection is made, what is the rate at which (a) the charge of the capacitor is increasing; (b) energy is being stored in the capacitor; (c)...
  14. Fat_Squirrel

    Circuits .. what is this question asking?

    Thanks heaps. Feeling stupid. Not good when I'm stuck on one of the first questions. Appreciate the quick response. Now I can enjoy my wine in peace.
  15. Fat_Squirrel

    Circuits .. what is this question asking?

    That makes sense. I assume I'm okay on the other two?
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