Recent content by the-alchemist

  1. T

    Conducting Spheres in Electrostatic Equilibrium

    Homework Statement Homework Equations potential, V = kq/r where k is 1/4 pi epsilon conservation of charge, sum of q = Q The Attempt at a Solution what i did is let q1 + q2 + q3 = 4*10^-4 then let V1 = V2 = V3 and solve for q3 but i end up getting q3 = 0.0002 which is...
  2. T

    Calculate voltage drop for a circuit

    Homework Statement Homework Equations V=IR (Ohm's Law) The Attempt at a Solution Now I'm able to get V=0 for when R=0 but I'm not sure as to how to solve for I when R=0. Can I ignore the 10 Ohm resistor as it's a short circuit as R=0.
  3. T

    Nortons Equivalent Circuit to find I

    Homework Statement Use Norton's Equivalent Circuit to find current I through 15ohm resistor. Homework Equations norton's theorem states that any two terminal network that includes many sources and resistors can be converted into an equivalent two terminal network with one current...
  4. T

    Calculating Resistance in a Series Circuit

    This really helped me! Thanks! I think I got it now. Correct me if I'm wrong, the current for the 20 ohm resistor is 1A and voltage of 20V. The 20 ohm and 10 ohm resistors are parallel thus they share the same voltage drop. Using the voltage divider principle I was able to get 10 ohms...
  5. T

    Calculating Resistance in a Series Circuit

    I've tried this method as well but I get stuck after calculating the voltage for the 20 ohm resistor. Correct me if I'm wrong but I thought any resistor that is parallel to an ideal voltage source has no effect on the voltage or current in the rest of the circuit. I'm still unable to get the...
  6. T

    Calculating Resistance in a Series Circuit

    Homework Statement I have been stuck at the following question for some time now. All help is appreciated. Homework Equations P=IV, V=IRThe Attempt at a Solution I have resolved the right side of the circuit into a 10 ohm resistor. I have also found the current in the 20 ohm resistor (0.4A). I...