In a Wheatstone-s bridge- there resistances mathrm-P- O and mathrm-R- connected in the three arms and the fourth arm is formed by two resistances mathrm-S-1- and mathrm-S-2- connected in parallel- The condition bridge to be balanced will be displaystyle dfrac-mathrm-P-mathrm-O-dfrac-2mathrm-R-mathrm-S-1-mathrm-S-2- displaystyle dfrac-mathrm-P-mathrm-O-dfrac-mathrm-R-mathrm-S-1-mathrm-S-2-mathrm-S-1-mathrm-S-2- displaystyle dfrac-mathrm-P-mathrm-O-dfrac-mathrm-R-mathrm-S-1-mathrm-S-2-2mathrm-S-1-mathrm-S-2- displaystyle dfrac-mathrm-P-mathrm-O-dfrac-mathrm-R-mathrm-S-1-mathrm-S-2-

Question

Toppr Admin · Accepted Answer

Equivalent resistance of the resistors S1 and S2 connected in parallel -xA0- -xA0- -xA0-1S-1S1-1S2-x27F9- -xA0- -xA0-S-S1S2S1-S2For balanced bridge- -xA0- -xA0-PO-RS -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-x27F9-PO-R-S1-S2-S1S2

In a Wheatstone's bridge, there resistances P, O and R connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for bridge to be balanced will be PO=2RS1+S2 PO=R(S1+S2)S1S2 PO=R(S1+S2)2S1S2 PO=RS1+S2

Equivalent resistance of the resistors S1 and S2 connected in parallel 1S=1S1+1S2⟹ S=S1S2S1+S2For balanced bridge, PO=RS ⟹PO=R(S1+S2)S1S2

Equivalent resistance of the resistors $S_{1}$ and $S_{2}$ connected in parallel $\frac{1}{S} = \frac{1}{S_{1}} + \frac{1}{S_{2}}$
$⟹$ $S = \frac{S_{1} S_{2}}{S_{1} + S_{2}}$
For balanced bridge, $\frac{P}{O} = \frac{R}{S}$ $⟹ \frac{P}{O} = \frac{R (S_{1} + S_{2})}{S_{1} S_{2}}$