Engineering Calculating current and voltage of a rectifier circuit

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The discussion revolves around calculating the average and RMS values of load current and voltage in a half-wave rectifier circuit using a step-down transformer with a 10:1 turns ratio and 230 V input. Initial calculations yielded incorrect values for RMS voltage and current, prompting a reevaluation of the approach. Participants emphasized the importance of sketching waveforms to visualize the circuit behavior and suggested modeling the circuit as an ideal transformer while considering secondary resistance. A voltage divider approach was recommended to accurately compute the half-rectified output. The conversation highlights the need for careful analysis and correct application of principles in electronics problems.
mooncrater
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Homework Statement


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".

Homework Equations


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

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 realized 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 ?
 
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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.
 
mooncrater said:
so, where am I wrong ?
.
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) 230√[/size]2 V
 
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mooncrater said:

Homework Statement


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".
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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