Dear All,
Its a simple electronics question but I cant get.
See image of a bridge here

http://www.facstaff.bucknell.edu/mastascu/elessonsHTML/Resist/Circuit2A01.gif

Lets say Ro=0 Ra = 1K, Rb=2K Rc=3k and Rs = 6600 ohms.

How do I calculate the Vout?

Question

Dear All,
Its a simple electronics question but I cant get.
See image of a bridge here

 http://www.facstaff.bucknell.edu/mastascu/elessonsHTML/Resist/Circuit2A01.gif

Lets say Ro=0 Ra = 1K, Rb=2K Rc=3k and Rs = 6600 ohms. 

How do I calculate the Vout?

anonymous · Answer

First solve the network of resistors Ra, Rb , Rc and Rs. Ra and Rb are in series. These make equivalent 
Rab=Ra+Rb, 
similarly    Rcs=Rc +Rs. Now Rab and Rcs are in parallel. Their equivalent Resistance is In series with Ro. Thus u will get a network of two resistors Ro and Req, and Vc. |dw:1368354460753:dw|. Veq is voltage around Req. Simply apply voltae divider around it. This can be shown as :  Now there are tow voltage divider networks. Voltage across Ra is 
$$V(Ra)= Ra*Veq/(Ra+Rb)$$. Similarly Voltage across Rc is $$V(Rc)=Rc*Veq/(Rc+Rs)$$.