Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 8 M

Year: MAY 2012

Let us mark the nodes first

Applying KUL at node a

I1 = I2 + I5

We know that I5 = 10A

$∴ 240-VA/3 = VA – Vo/6 + 10 \\ 6(240 –VA) = 3 (VA-Vo) + (3*6*10) \\ ∴ 1440 -6VA = 3VA – 3VB + 180 \\ ∴ 9VA – 3VB = 1260 \\ ∴ 3VA- VB = 420$

Applying KCL at node b

I2= I3 +I4

$∴ VA-VB/6 = 2VB + 5 (VB –VC)/60 \\ ∴ 60 VA- 60VB= 12VB+ 30 VB- 30VC$

But, VC= 60V

$60VA- 102VB = -30*60 = -1800 \\ 60VA- 100VB = -1800$

Solving eq. 1 & 2 simultaneously we get

VA= 181.46 V

BB= 124.4 V

$∴ V1= 240 – VA \\ = 240 – 181.46 \\ = 58.54 V \\ V2= \sqrt{(B-VC)} \\ = 124.4- 60 \\ = 64.4 V$

Power absorbed by the 6Ω resistance

$P6= (VA- VB) 2/ 6 \\ = (181.46 – 124.4)2/6$

∴ P= 30963 Watt