In fig given above, the parallel combination of resistance of 400 Ω and voltameter (V) of resistance 400 Ω connected across it. The equivalent resistance is 200 Ω. Then in given circuit is equivalent.

(i) Applying Kirchhoff's second law in closed loop OPLMN

    100i₂ - 1OOi₁ + 100i₃ = 0

     i₂ - i ₁+ i₃ = 0

⇒ - i₁ + i₂ + i ₃ = 0

or         - I    (i ₁- i₂-i₃) = 0         ...(1)

(ii) In closed loop (ii) OPQRLPO, we have Kirchhoff' s second law,

     lOOi ₁+ (i₁ + i ₃)200 = 10

or    100i₁ + 200i₁ + 200i₃ = 10

⇒              300i ₁+ 200i₃ = 10

            3i₁ + 2i₃ = 0.1                ...(2)

(iii) In closed loop (iii), MTSRLM, according Kirchhoff s second law

             (i₂ -i₃) 200 - (i₁ + i₃)200 - lOOi₃ = 0

or        200i₂ - 200i₃ - 200i₁ - 200i₃ - 100i₃ = 0

                             200i₂ - 200i₁ - 500i ₅ = 0

or                         -200i₁ + 200i₂ - 500i₃ = 0

or                             i₁-i₂ + 2.5i₃ =0                   ...(3)

From eqn. (1) and (3), we get

           (i ₁- i₂ - i₃ ) - (i ₁- i₂ + 2.5i₃ )=0

⇒                i ₁- i₂ - i₃ -i ₁+ i ₂- 2.5i₃ = 0

                                          -3.5 i₃ = 0

or                             i₃ = 0                            ...(4)

Putting value i₃ in eqn. (2), we get

3i₁ + 2 x 0 = 0.1

            3i ₁ =0.1