(Va-30)/82 + (Va-40)/68 + Va/147 = 0 // summation i out = 0
multipy all by 409836
4998Va - 149940 + 6027Va - 241080 + 2788Va = 0
Solving Va = 28.308116V

I believe it is a mathematics error not conceptual error.
BTW, I am not trying to be mean but your tutor shouldnt be giving much big resistor number that are difficult to calculate. My tutorials answers are much more beautiful.

 

hitmen wrote:

(Va-30)/82 + (Va-40)/68 + Va/147 = 0 // summation i out = 0
multipy all by 409836
4998Va - 149940 + 6027Va - 241080 + 2788Va = 0
Solving Va = 28.308116V

I believe it is a mathematics error not conceptual error.
BTW, I am not trying to be mean but your tutor shouldnt be giving much big resistor number that are difficult to calculate. My tutorials answers are much more beautiful.
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hitmen You do not need to multiply by 409836; or any proportional term to solve for Va. Simply solve it directly! Let me show you (I will show every step for clarity):

(Va-30)/82 + (Va-40)/68 + Va/147 = 0

Expand the above equation:

Va/82 - 30/82 + Va/68 - 40/68 + Va/147=0

Factor out Va:

Va(1/82 + 1/68 + 1/147) - 30/82 -40/68 =0

Could add the fractions by finding a common denominator-but since using a calculator will change all fraction values to decimal form:

Va(0.012195 + 0.014705 + 0.006802) - 0.365853 - 0.588235 = 0

Va(0.337037) - 0.954088 = 0

Va(0.337037) = 0.954088

Va = 0.954088/0.337037

Va = 28.308116 Volts

This is the same answer you got hitmen without multiplying through with a constant!

hitmen solving for any variable after multiplying through with a proportional term in any linear equation being equal to zero or not; will not change the final answer in the variable-since all you are doing is changing both sides of the equation in proportion.

 

Here is the Millman's Theorem solution for anyone interested.

\(V_a\ =\LARGE\ \frac{\frac{30}{82}\ +\ \frac{40}{68}\ +\ \frac{0}{(47+100)}}{\frac{1}{82}\ +\ \frac{1}{68}\ +\ \frac{1}{(47+100)}}\)

\(V_a\ =\ 28.308116\ Volts\)

hgmjr