In a Wheatstone's bridge, three resistances P, Q and R are connected in the three arms and fourth arm is formed by two resistances S₁, and S₂, connected in parallel. The condition for the bridge to be balanced will be

• A

  $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{2\mathrm{R}}{\left({\mathrm{S}}_1+{\mathrm{S}}_2\right)}$

• B

  $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left({\mathrm{S}}_1+{\mathrm{S}}_2\right)}{{\mathrm{S}}_1{\mathrm{S}}_2}$

• C

  $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left({\mathrm{S}}_1+{\mathrm{S}}_2\right)}{2{\mathrm{S}}_1{\mathrm{S}}_2}$

• D

  $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\left({\mathrm{S}}_1+{\mathrm{S}}_2\right)}$