Apply KCL at node a-

$\dfrac{V_a-240}{3}+$  $\dfrac{V_a-V_b}{6 } +10 = 0$

$3V_a - V_b = 420 \cdots(1)$    

Apply KCL at node b-

$\dfrac{V_b-V_a}{6}$ $+ \dfrac{V_b}{30}$ $+ \dfrac{V_b-V_c}{12} = 0$

$-10V_a + 17V_b - 5V_c = 0 \cdots (2)$

• Node c is directly connected to a voltage source of 60 v. Hence, we can write voltage equation at Node c.

$V_c = 60V$

• On solving equation [I] and [II], we get-

$V_a = 181.46 V$

$V_b = 124.39 V$

• Thus,

$V_1 = V_a - 240 = - 58.54 V$

$V_2 = V_b - V_c = 64.39 V$

• Now,
• Current through 6 ohm resistor$=\dfrac{V_a-V_b}{6} = 9.51A$
• Therefore, Power absorbed by 6 ohm resistor = I2R = (9.51)2(6) = 542.64 W