How do diodes impact voltage measurements in a full-wave rectifier circuit?

[TsAmE]

Homework Statement

This circuit has kind of been confusing me. In my notes it says V across capacitor = V - 2V, due to the voltage drop across the 2 diodes per half cycle (assuming that the diodes used have a 1V drop each). But (refering to the attachment with the red line), for the half cycle starting at the top diode, by the time you reach the top of the capacitor, wouldn't the voltage have only traveled through the first diode, thus V across the capacitor = V - 1V?).

Homework Equations

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The Attempt at a Solution

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[berkeman]

Homework Statement

This circuit has kind of been confusing me. In my notes it says V across capacitor = V - 2V, due to the voltage drop across the 2 diodes per half cycle (assuming that the diodes used have a 1V drop each). But (refering to the attachment with the red line), for the half cycle starting at the top diode, by the time you reach the top of the capacitor, wouldn't the voltage have only traveled through the first diode, thus V across the capacitor = V - 1V?).

Homework Equations

n/a

The Attempt at a Solution

n/a

You've only highlighted half of the current path. The current flows through 2 diodes in the loop from the source to the storage cap. Which 2 diodes depends on whether the source polarity is positive or negative.

 

[Zryn]

What you are saying would be true if you don't take into account the idea that current will not flow without a closed loop.

So in order for there to be a closed loop and for current to flow, the second diode in the return path must be turned on.

Wikipedia has a decent introduction for Diode Bridge, with diagrams.

http://en.wikipedia.org/wiki/Diode_bridge