Results 1 to 2 of 2

Thread: Voltage drop

  1. #1 Voltage drop 
    Forum Freshman
    Join Date
    Nov 2006
    A 6V DC voltage source is connected across both a 5 Kohm resistor and a 10nF capacitor in series. The voltage drop across the resistor is equal to what

    Reply With Quote  


  3. #2  
    Forum Masters Degree bit4bit's Avatar
    Join Date
    Jul 2007
    The expression for the voltage across the capacitor when charging, is almost the same as when it is discharging (as in your other question), except the time will not have the negative co-efficient. This is due to the application of Kirchoff's second law in the opposite direction.

    The expression is therefore:

    V_c = V_0 *e^(t/RC)

    (this expression can be used alongside the expression Q = Q_0 *e^(t/RC) since V = Q/C )

    For a DC supply, the sum of the voltage drops of ALL components in the circuit (including internal resistance of the battery) is equal to the EMF of the supply.

    That is, in this case:

    V_R = V_supply - V_c

    Reply With Quote  

Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts