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!

Homework Help: Capactive network

  1. Sep 16, 2006 #1
    Hi all,
    I am new to this group and joined after reading this intresting discussion. In extention th the David, I've one doubt to find the voltage between two capa. Suppose a capa network is like Vsupply(1V)-C1-C2-gnd; Let say Vx is the between node name. Vx1=C1/(C1+C2)*1v =0.5v if C1=C2; Now Vsupply=2v, I want to calculate new Vx1, say Vx2, from charge-conservation theory.Previous total charge was 1v*[C1*C2/(C1+C2)] -series combination; Current total charge is (2v-Vx2)C1 +Vx2*C2; If you equate both charge, Vx2=1/(C2-C1)*[1v*{C1*C2/(C1+C2)}-2v*C1] which infinity if C1=C2.
    Pls you (Russ or anybody) point out where I am wrong; Russ
     
  2. jcsd
  3. Sep 16, 2006 #2

    Astronuc

    User Avatar
    Staff Emeritus
    Science Advisor

    One cannot equate both charges. If one doubles the voltage, the charge on the capacitors changes, in fact it doubles, and the charge on each capacitor doubles and is proportionally distributed as before.

    Where there are two elements in series, there is a voltage drop across each.

    Given a V --| C1 |-----| C2 |----- gnd, V = V1 + V2, but capacitors in series are treated like resistors in parallel by virtue of the relationship between charge and voltage. Current is a transient property in capacitors, since when a voltage is applied, current flows until a new equilibrium voltage (charge) is achieved.

    The combined capacitance of C1 and C2 is given by,

    1/C = 1/C1 + 1/C2, or C = (C1*C2)/(C1+C2). If C1 = C2, then C = (C1*C1)/(C1+C1) = (C1)/2

    And V = Q/C => V = Q (C1+C2)/(C1*C2), and by charge conservation, the charge is the same on each capacitor, and the net charge in a capacitor is zero, i.e. the negative charge on one plate is balanced by the same magnitude of positive charge on the other plate.

    So Q = Q1 = Q2 or C*V = C1*V1 = C2*V2, so one can find V1 or V2 in terms of V,

    V1 = (C*V)/C1 = [(C1*C2)/(C1+C2) / C1] V = C2/(C1+C2) V, and likewise
    V2 = C1/(C1+C2) V.

    When the voltage is change, the charge on each capacitor changes simultaneously.

    See - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html
     
    Last edited: Sep 16, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...
Similar Threads for Capactive network Date
Capacitor network Nov 11, 2017
Asymptotic magnitude Bode plot for a network function Apr 19, 2017
Network function for an opamp circuit Mar 13, 2017