Let the conductance be represented by g. Let all the sources be current sources. For this, a voltage-source in series with a resistor is transformed into its equivalent current source. This is done in Fig..
Observing the circuit, g11 = (1/5) + 0.6 = 0.8, g22 = 0.40 + 0.2 = 0.6 g12 = 0.2,
Current sources : + 5 amp into ‘A’ + 5.67 amp into ‘B’
VA = 4.134/0.44 = 9.4 volts,
VB = 5.536/0.44 = 12.6 volts.
Current in 5-ohm resistor = (VB −VA)/5 = 0.64 amp
Check : Apply Thevenin’s Theorem :
VA = 10 × (10/12) = 8.333 V
VB = (17/3) × 2.5 = 14.167 V
VTH = 14.167 − 8.333 = 5.834 V
RTH = 4.167 I5 = 5.834/(4.167 + 5) = 0.64 A
