How does voltage splits when 2 devices was attached together
Name lights A &amp; B. 
Do you have links or technical descriptions of A &amp; B? 

The following describes ways to determine if your power supply is overloaded.  

Note that an overloaded power supply may work indefinitely, or fail quietly or burst into flames. 

Presumably both are 12V rated.
They will also have a
Power (P) rating and a 
Current (I) rating.   

Approximately P = V x I.
And I = P/V. 

Va = voltage rating of A, IA = current rating of A etc.
It is essential that either:


Va = Vb = Vpsu
(Ia + Ib) &lt;= Ipsu
LEDs are operated in parallel from psu.
ie connections are   


V+_psu = V+_A = V+_B  
V-_PSU = V-_A = V-_B



or  


A very special case
(Va + VB) = Vpsu
AND Ia = Ib &lt;= Ipsu
LEDs are run in series from psu.
Connections are:   


V+_psu = V+_A  
V-_A = V+_B  
V-_B = V-_psu
This is unusual and best avoided. 



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Test:
As above: If Ia + Ib > Ipsu the psu is overloaded. 

Operate A only from psu (power supply unit). Note brightness.
Operate B only from psu. Note brightness.
Operate A&amp;B together from psu.
If brightness of A is lower than when by itself and/or brightness of B is lower than by itself then psu is overloaded. 

If you have a DMM ()Digital Multimeter) - which is a really useful tool to have, then you can check as follows.

Set meter to 20 VDC or higher range.
Measure unloaded psu Vout = Voc
Voc probably 12V - 15V depending on psu.

Load with  A.
Measure Vo  = Vo_A
Vo_A probably 10V - 13V

Load with  B.
Measure Vo  = Vo_B
Vo_B probably 10V - 13V

Load with A &amp; B. 
Measure Vo = Vo_AB
If Vab is well below Va or Vb and/or well below 12V then psu is overloaded.