Electrical Engineering - Branch, Loop and Node Analyses - Discussion
Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 1)
1.
Find the node voltage VA.
Discussion:
39 comments Page 1 of 4.
Pewds said:
2 years ago
@All.
Here is the soltion.
Millman's Theorem.
(Σ E/R)/Σ 1/R.
Vab = (12/49 + 0/24 +6/80) / (1/49 + 1/24 + 1/80).
Vab = 4.2896 V.
Here is the soltion.
Millman's Theorem.
(Σ E/R)/Σ 1/R.
Vab = (12/49 + 0/24 +6/80) / (1/49 + 1/24 + 1/80).
Vab = 4.2896 V.
(9)
Sharathkumar said:
3 years ago
((Va-12)/49) + (Va/24) + (Va-6)/80))=0
((24Va-288 + 49Va)/1176) + (Va-6)/80))=0
1920Va-2340 + 3920Va + 1176Va-7056=0
7016Va-30096 =0,
7016Va = 30096,
Va = 30096/7016,
= 4.28.
((24Va-288 + 49Va)/1176) + (Va-6)/80))=0
1920Va-2340 + 3920Va + 1176Va-7056=0
7016Va-30096 =0,
7016Va = 30096,
Va = 30096/7016,
= 4.28.
(9)
Nami-Swan Jyv said:
3 years ago
KCL: Iin = Iout.
I1 + I3 = I2.
I1 - I2 + I3 = 0.
Nodal Analysis:
(Start node minus End node)/Resistor.
((V1- Va)/R1) - ((Va-0)/R2) +((V2-Va)/R3) = 0.
Shift Solve;
Va= 4.28v.
or
Va = 4.25v.
I1 + I3 = I2.
I1 - I2 + I3 = 0.
Nodal Analysis:
(Start node minus End node)/Resistor.
((V1- Va)/R1) - ((Va-0)/R2) +((V2-Va)/R3) = 0.
Shift Solve;
Va= 4.28v.
or
Va = 4.25v.
(5)
Vishal P said:
7 years ago
It's simple.
According to the KCL summation of all the currents in the circuit is zero. That only is applied here.
According to the KCL summation of all the currents in the circuit is zero. That only is applied here.
(2)
Abyssinia said:
4 years ago
Thanks all for explaining.
(1)
Adel saeed said:
5 years ago
Thanks @Abdullah.
(1)
Balu said:
5 years ago
Thanks all for giving the solution.
(1)
Yamanappa said:
6 years ago
((Vs1 - Va) /R1)+((Vs2 - Va)/R3)) = Va/R2.
(1)
Rakesh. Parida said:
5 years ago
Thanks all for explaining it.
Dinesh sah said:
6 years ago
4.28 is the correct answer.
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