# Calculating current and voltage of a rectifier circuit

1. Oct 18, 2015

### mooncrater

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. Oct 19, 2015

### Bernie G

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.

3. Oct 21, 2015

### 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

4. Oct 21, 2015

### rude man

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.