Question about full wave bridge rectifier Sorry if this is a stupid question, what I know is when a diode is forward biased, it conducts current, when it is reverse biased, it acts as an open branch if its $ PIV \geq V_{in}$ 
But in this picture:

At the node between $D1$ and $D3$ ,$D1$ is forward biased and $D3$ is reverse biased, so that's fine, now at the node between $D2$ and $D3$ why does the current pass through $D2$ and not $D3$? Isn't $D3$ supposed to be forward biased at that junction too? Thanks.
 A: The node between D2 and D3 is grounded, so its potential is 0 V. The node between D3 and D1, as you stated, has positive potential, so the current can't go through D3.
On the other hand the node between D2 and D4 has a negative potential, so the current can go in that direction, and indeed it does.
A: 
what I know is when a diode is forward biased, it conducts current,
  when it is reverse biased, it acts as an open branch

There's a simpler, clearer way to think about it - a diode (ideally) allows current through in just one direction.
So, it's easy to see that current entering the top node of the bridge can only exit through through D1 since there can be no current (ideally) 'backwards' through D3.
However, for the current entering the left node of the bridge, there could be a current exiting through D3 but only if the cathode is negative with respect to the anode.  But we know this isn't the case, i.e., the cathode is positive with respect to the anode while, for D2, the cathode is negative with respect the anode.
Thus, the current is 'downhill' through D2 and not 'uphill' through D3. 
A: Assume that $V_{\rm X}=0 \,\rm V$ and noting the labels (signs) on the secondary of the transformer.
. 
$V_{\rm X}>V_{\rm W}$ so diode D1 is forward biased and conducting.  
$V_{\rm Y}>V_{\rm Z}$ so diode D2 is forward biased and conducting.  
$V_{\rm Y}>V_{\rm X}$ so diode D3 is reverse biased and not conducting.  
$V_{\rm Z}>V_{\rm W}$ so diode D4 is reverse biased and not conducting.   
Assuming that the potential drop across the conducting diodes is very small nodes $X$ and $W$ are approximately at zero potential whereas nodes $Y$ and $Z$ are at a higher potential.
