electric circuits

Discussion in 'Homework Help' started by srinivasaphanikiran, Apr 22, 2011.

  1. srinivasaphanikiran

    Thread Starter New Member

    Apr 22, 2011
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    when two star circits are connected with all the values(impedences) are equal the current in the nuetral is "0". But in the same situation two ckts A and B of R Y B phases on each circuits each having three capacitors of capacitence C with fuses is connected in parallel.If one fuse blows on ckt A for one PHASE is blown means one capacitor in three parallel capacitors is open.ckt B has no change what will be the current in the nuetral wire if ammeter is connected.Voltage V and Frequency 60HZ?

    Please help me in this question
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If I understand correctly, you are asking what happens when the load capacitance on one phase is reduced the half that value on the other two phases...

    If the phase with reduced capacitance has a capacitance load of C and the other two phases have a value of 2*C then ...

    The rms current magnitude in the neutral would be ω*C*Vrms amperes. Vrms is the line-to-neutral voltage.
     
  3. srinivasaphanikiran

    Thread Starter New Member

    Apr 22, 2011
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    0
    iam sorry may be i dint explained u correctly it was more related to symmetric components and two star ckts were connected not single one and in each phase THREE CAPACITORS with c capacitence was connected in parallel with fuses and in this condition as two bridges were in balaced condition Nuetral current is 0.But if one fuse blown in R phase(R phase capacitence is 2c instead of 3C as they are in parallel) in ckt A and in ckt B 3 capacitors are in parallel but in Y and B on two bridges it was same(3c is capacitence) so current will flow in nuetral. how much current if ammeter is connected. THIS WAS VERY GOOD AND TYPICAL IT WAS ASKED IN GATE(IIT TEST) IN INDIA .I THINK U GOT THE QUESTION
     
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