What about the load resistor? Not taking into consideration? Since the voltage will pass through the load and there will be a drop in current..voltage as well since back to ground.
Why do you divide the output peak voltage by 2/pi? isn't the output peak voltage what you measure with the multimeter? isn't it the same thing as the output voltage?
Am getting 14v a.c. when dc its 11.60 what could be the problem am trying to repair a batt charger am a mechanic not elect just trying to diy and learn elect
You either divide by 0.707 or multiply by square root of 2 (basically the same thing) because you cannot possibly operate on the peak voltage. RMS value is the "effective" voltage that is best comparable to pure DC voltage.
You are all ibiots. There are no diodes with 0.7 voltage drop anymore! Already 10 years ago Rectifying/Blocking didoes are 0.4v and shotky diodes 0.2v@@zahoorahmad9132
My goodness. He took 9 minutes and 26 seconds to say what could have been said in 26 seconds. And if you are going to teach to the whole world, perhaps you should stick to your native tongue. Strong accents are hard to follow, and it would be even more so for a person for whom English is a second language. You have learned your electronics lessons, as well you should if you teach. Just as much, you should take some English diction lessons if you plan to teach in English.
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Thank you for the clarity. You made it look so easy
Thank you Trevor! This was vary helpful and easy to follow. Appreciate it.
Great explanation!! Love from Bangladesh ❤️
thank you so much!!! This helped me a lot
Nicely done ✔️
Thanks Salman !
What about the load resistor? Not taking into consideration? Since the voltage will pass through the load and there will be a drop in current..voltage as well since back to ground.
Why V peak out where we get 1.4v?
But when we filter 12 volt ac it becomes almost 16 volt dc. Can we connect a proper 12 volt device to this 16 volt dc output?
he just said it is 9.9.... so which is it?
(this is like brain surgery after all...)
From where u bring 1,4v
Why do you divide the output peak voltage by 2/pi? isn't the output peak voltage what you measure with the multimeter? isn't it the same thing as the output voltage?
Am getting 14v a.c. when dc its 11.60 what could be the problem am trying to repair a batt charger am a mechanic not elect just trying to diy and learn elect
why we want to divide 12v by 0.707?
It is to calculate the RMS value
You either divide by 0.707 or multiply by square root of 2 (basically the same thing) because you cannot possibly operate on the peak voltage. RMS value is the "effective" voltage that is best comparable to pure DC voltage.
You can do it by (root 2)*12. All the same.
thnx@@ironh2017
What is rms ?
Root mean square value (Vrms= peak value ×0.707)
Where did pi come from?
Pi results from the mathematical process of integration, or summing.
1.4 volt where is.
Voltage drop across each diode=0.7V: During each Half cycle 2 Diodes conduct so total voltage drop across 2 diodes during each half cycle=2X0.7=1.4V
@@zahoorahmad9132 thanks!
You are all ibiots. There are no diodes with 0.7 voltage drop anymore!
Already 10 years ago Rectifying/Blocking didoes are 0.4v and shotky diodes 0.2v@@zahoorahmad9132
My goodness. He took 9 minutes and 26 seconds to say what could have been said in 26 seconds. And if you are going to teach to the whole world, perhaps you should stick to your native tongue. Strong accents are hard to follow, and it would be even more so for a person for whom English is a second language. You have learned your electronics lessons, as well you should if you teach. Just as much, you should take some English diction lessons if you plan to teach in English.
😂😂yu need them acscent learning this is noothing try listening to an ghana