Question

Find the current I in the j10 Ω branch of the given circuit shown in Fig.  using the Nodal Method.

Answer 1

There are two principal nodes out of which node No. 2 has been taken as the reference node. 

V₁(1/(6 + j8) + 1/(6 - j8) + 1/j10) - (100 ∠ 0º)/(6 + j8) (100 ∠- 60º)/(6 - j8) = 0

V₁(0.06 −j0.08 + 0.06 + j0.08 −j0.1) = 6 −j8 + 9.93 − j1.2 = 18.4 ∠−30º 

∴ V₁(0.12 −j0.1) = 18.4∠30º or V₁ × 0.156 ∠−85.6º = 18.4 ∠− 30º

∴ V₁ = 18.4 ∠− 30º/0.156∠−85.6º = 118∠ 55.6ºV 

∴ V = V₁/j10 = 118∠55.6º/j10 = 11.8∠−34.4ºA