DC measuring

Alex18

New Member
Need help.
Which DC current can you measure according to picture a, b, c? I think its 19,8V because a diode needs 0,7V so 20,5V-0,7V=19,8V
 

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You would be measuring DC voltage [gleichspannung], not current.
(Voltage is equivalent to pressure, current is equivalent to flow).

The capacitor would charge to the peak voltage, 1.414 x the RMS rating (minus the 0.6 - 0.7V diode drop).

If the AC voltage is 20.5V RMS, the peak will be 20.5 x 1.414 ~= 29V
- 0.6 = 28.4V

In the first circuit A will be positive, in the other two B is positive.
 
20,5VeFF is in a foreign language so we do not know if it is RMS or is peak-to-peak.
Why is the transformer measured in V but the voltage across the capacitor is measured in U?
 
20,5VeFF is in a foreign language so we do not know if it is RMS or is peak-to-peak.
Why is the transformer measured in V but the voltage across the capacitor is measured in U?

Presumably it's obviously German? - and also presumably V is AC and U is DC?.

As you're well aware, mains transformers are always specified in RMS.
 
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