Finding Turns Ratio and Average Current in the Primary Winding

[JoeMarsh2017]

Homework Statement

A half wave rectifier circuit has a transformer inserted between the source and the remainder of the circuit. The source is 240Vrms and 60 Hz, and the load resistance is 20ohms.
1) Determine the required Turns Ratio of the transformer such that the average load current is 12A?
2) Determine the the average current in the primary winding of the transformer.

So. we have the "source" at 240 Vrms / 60 hz feeding a transformer to a Diode(assumed to be ideal with no voltage drop) and a 20ohm resistor.

The Turns ratio is basic, which is V primary/V secondary... but I already know that the secondary has 12 Amps across the 20 Ohm load resistor.

Homework Equations

This could go in either RE or Solution so I will put it here:

to find Average Load Current, we would normally take the Peak Voltage/ Resistance*Pi
So my next planned move was to take 12A times(20*3.14) which is 12A times 62.8 ohms = 753.6 Volts

My turns ratio is Vp/Vs so 240/754 = 0.32 A BUT the Primary side is in Vrms...so that needs to be converted

The Attempt at a Solution

Can some one set me straight...its been a semester since I took my Motors/Transformer class... I believe once I know the current and Voltage in the secondary...and the Voltage in the Primary...I can work it backwards to the primary... Thanks for your help in advance! JM

 

[JoeMarsh2017]

Oh man...I just found what I think I was missing..since I know the power in the secondary, 12A times 754 Volts...I get 9,048 Watts 9,048 divided by 240 Volts on the primary would give me the Current in the Primary...right? I am still getting thrown for a loop here because the Resistance is not at 20 ohms...its at 62.8 due to the (20 ohms x Pi = 62.8)

Am I right...or am I using the wrong Voltage on the secondary?

 

[CWatters]

12*754 would be the average current times the peak voltage so that's not correct.

If the peak voltage after the diode is 754V then what is the peak to peak voltage before the diode?

 

[JoeMarsh2017]

OK CWatters ... I beielve I misundestood the relationship between the Iavg, Irms and Ipeak current

 

[JoeMarsh2017]

Re-doing this post since I am starting to figure it out...

754 Volts is the Vmax I can take 75/Sqr rt 2 which will give me my rms Volts =532.88

 

[JoeMarsh2017]

OK the rectifier is ideal, the loss is no loss (voltage is high enough we can we neglect it...
The voltage after the rectifier is DC Voltage

 

[CWatters]

It's not DC after the diode. There is no mention of a smoothing capacitor.

After the rectifier you will have a half sine wave with Vm=754V. So before the rectifier you will have a full sine wave with a peak to peak voltage of double that.

What's the peak to peak voltage on the primary if the rms on the primary is 240V?

Now you have peak to peak voltages for both sides and can work out the ratio.

 

[JoeMarsh2017]

240rms X sqr rt2 =339 volts

 

[JoeMarsh2017]

Now you have peak to peak voltages for both sides and can work out the ratio

so, since we know that sqr rt2(532 Vrms)/20*pi = 12A which is our average current

I would compare 532vrms/240 vrms to get my turns ratio..its basically a 1:2.2 step up transformer

 

[CWatters]

I got the same answer by a similar method...

After the diode the voltage is 754V peak.
So before the diode (eg the secondary voltage) is 1508 V peak to peak.
The Input voltage is 240 Vrms so the Input peak to peak voltage is 240/0.3535 = 679V

So it's a step up transformer with ratio 1 to 1508/679 = 1:2.2

 

[rude man]

"Average" current is not the same as "rms" current. Has that been factored in on the many posts in this thread?

 

[CWatters]

Good catch. I assumed they meant rms when they said the current was 12A average.

 

[rude man]

Good catch. I assumed they meant rms when they said the current was 12A average.

Good chance they did!