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Calculating current and voltage of a rectifier circuit

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    The question is :
    "A step down transformer having turns ratio 10:1 and input 230 V,50 Hz is used in a half wave rectifier. The diode forward resistance is 15 ohms and resistance of secondary winding is 10 ohms . For a load resistance of 4 kilo ohms, calculate the average and R.M.S values of load current and voltage".

    2. Relevant equations
    $$I_{R.M.S}=I_{MAX}/2$$
    $$I_{avg}=I_{MAX}/pi$$---(For a half-wave rectifier)
    $$I_{R.M.S}=V_{R.M.S}/R_{Total}$$
    $$E_{s,R.M.S}/E_{p,R.M.S}=N_2/N_1$$



    3. The attempt at a solution
    So, using the ##4^{th}## equation :
    ##E_{s,R.M.S}/230=1/10##
    So,
    $$E_{s,R.M.S}=23 V$$
    Because of which,
    $$ I_{R.M.S}=23/(4025) A$$
    So , ##I_{R.M.S}=5.71 mA##
    But even looking at the answer at this point of time , I realised that both of these answers were wrong.
    The given value of ##E_{R.M.S}=16.1624 V## and ##I_{R.M.S}=4.0406 mA ##
    so, where am I wrong ?
     
  2. jcsd
  3. Oct 19, 2015 #2
    The diode forward resistance is 15 ohms half the time and infinite resistance half the time, so the load RMS voltage is half your value.
     
  4. Oct 21, 2015 #3

    NascentOxygen

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    Staff: Mentor

    The first place where you went wrong is to be attempting this problem without first sketching a large, clear diagram of the waveforms under investigation. A diagram is an essential aid to electronics problems.

    Also, remember that a mains voltage of 230V is a sinusoid of peak value (i.e., max) 2302 V
     
  5. Oct 21, 2015 #4

    rude man

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    Homework Helper
    Gold Member

    Model the circuit as an ideal transformer but with the given secondary resistance. I guess primary resistance is assumed zero which is unrealistic but then so are most textbook problems.

    So form a voltage divider between the transformer secondary resistance, the diode resistance and the 4K load. The compute the half-rectified output voltage and current.

    Most of the equations you cite look of questionable relevance. Try to work from math principles instead.
     
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