Wheatstone Bridge

For a certain adjustment of Q, V_BD = 0, therefore no current flows through the galvanometer.  

→     V_B = V_D or V_AB = V_AD  

→    l1.P = l2.R  

→

                            Alt txt :   wheatstone bridge

Likewise, V_BC = V_DC Þ l₁.Q = l₂.S  

Dividing (1) by (2), we get, P / Q = R / S  

Illustration:  

What’s the effective resistance of the circuits shown in the adjacent figure?

Alt txt : effective resistance of the circuits shown

Solution:  

(a)    It is a Wheatstone bridge that is balanced. Hence, the central resistance labeled ‘C’ can be assumed as ineffective.

        →     R_eq = R.

(b)    The resistance R is in parallel with a balanced Wheatstone bridge.

        →    R_eq = R.R / R + R = R / 2