Capacitor transient response - please help

Discussion in 'General Electronics Chat' started by e-akamuzja, Feb 6, 2009.

  1. e-akamuzja

    Thread Starter New Member

    Feb 6, 2009
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    Hi folks.
    I have my examinations soon and there's no one around who can help me.

    I stuck with Worksheet "Time constant calculations", Question 28, Figure 4: can't understand how to determine initial and final voltage for a capacitor when it is a part of a resistor series-parallel combination circuit. Why should the answer be V(C initial)=10V and V(C final)=12V???? :confused:

    Your help will be very much appreciated
     
  2. e-akamuzja

    Thread Starter New Member

    Feb 6, 2009
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  3. mayiru

    New Member

    Feb 6, 2009
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    0
    Its quite simple. For the 4th diagram, initially the switch is open, so the voltage across the capacitor is given by 12(5)(5+1) = 10 volts (Potential divider rule. Just remove the switch as it's open).

    When the switch is closed, it shorts the 1 ohm resistor, and the capacitor will charge towards 12V. So the final potential will be 12V.

    Potential at any time after closing will be given by V = 10(initial voltage) + 12(1-exp(-t/RC))
    Where R = 6 ohms, C = the capacitance, and t is the time at any instant in seconds.
     
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