In the network of Fig. determine the current flowing through the bra

1.	In the network of Fig. determine the current flowing through the branch of 4 Ω resistance using nodal analysis.
Answer» We will find voltages V_A and V_B by using Nodal analysis and then find the current through 4Ω resistor by dividing their difference by 4. V₂(1/5 + 1/4 + 1/j2) - V_B/4 - (50∠30º)/5 = 0 ∴ V_A(9 −j10) − 5V_B = 200 ∠30º ...(i) Similarly, from node B, we have V_B(1/4 + 1/2 + 1/-j2) - V_A/4 - (50 ∠ 90º)/2 = 0 ∴ V_B (3 + j2) −VA = 100 ∠90º = j 100 ...(ii) VA can be eliminated by multiplying. Eq. (ii) by (9−j10) and adding the result. ∴ V_B(42 −j12) = 1173 + j1000 or V_B = (1541.4 ∠ 40.40º)/(43.68 ∠-15.9º) = 35.29∠56.3º = 19.58 + j29.36 Substituting this value of V_B in Eq. (ii), we get V_A = V_B(3 + j2) −j100 = (19.58 + j29.36) (3 + j2) −j100 = j27.26 ∴ V_A −V_B = j 27.26 − 19.58 −j29.36 = − 19.58 −j2.1 = 19.69∠186.12º ∴ I₂ = (VA −VB)/4 = 19.69∠186.12º/4 = 4.92∠186.12º

In the network of Fig. determine the current flowing through the branch of 4 Ω resistance using nodal analysis.

Answer»

We will find voltages V_A and V_B by using Nodal analysis and then find the current through 4Ω resistor by dividing their difference by 4.

V₂(1/5 + 1/4 + 1/j2) - V_B/4 - (50∠30º)/5 = 0

∴ V_A(9 −j10) − 5V_B = 200 ∠30º ...(i)

Similarly, from node B, we have

V_B(1/4 + 1/2 + 1/-j2) - V_A/4 - (50 ∠ 90º)/2 = 0

∴ V_B (3 + j2) −VA = 100 ∠90º = j 100 ...(ii)

VA can be eliminated by multiplying. Eq. (ii) by (9−j10) and adding the result.

∴ V_B(42 −j12) = 1173 + j1000

or V_B = (1541.4 ∠ 40.40º)/(43.68 ∠-15.9º) = 35.29∠56.3º = 19.58 + j29.36

Substituting this value of V_B in Eq. (ii), we get

V_A = V_B(3 + j2) −j100 = (19.58 + j29.36) (3 + j2) −j100 = j27.26

∴ V_A −V_B = j 27.26 − 19.58 −j29.36 = − 19.58 −j2.1 = 19.69∠186.12º