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 2 of 4.
PRIYANKA said:
8 years ago
According to Kirchofs current law.
Total incoming currents= total outgoing currents.
((12-VA) /49) + ( (6-VA) /80) = (VA/24).
VA=4.25.
Total incoming currents= total outgoing currents.
((12-VA) /49) + ( (6-VA) /80) = (VA/24).
VA=4.25.
Pewds said:
3 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)
Rj naveed said:
9 years ago
Node analysis can be solved by Kirchhoff's 1st law mean sum of all current toward junction or away from junction is equal to zero.
Krishna said:
1 decade ago
((va-12)/49)+(va/24)+((va-6)/80)=0
((va-12)(1920)+(va)(3920)+(va-6)(1176))/94080=0
va(7016)-30096=0
va=30096/7016
va=4.28
((va-12)(1920)+(va)(3920)+(va-6)(1176))/94080=0
va(7016)-30096=0
va=30096/7016
va=4.28
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)
Ashiwani kumar said:
1 decade ago
Applying nodal analyses at node A.
I1 = (I2+I3).
(12-Va)/49 = Va/24+(Va-6)/80.
(Va = 4.289).
I1 = (I2+I3).
(12-Va)/49 = Va/24+(Va-6)/80.
(Va = 4.289).
Sam said:
1 decade ago
(VA-12)/49+VA/24+(VA-6)/80 = 0.
0.02(VA-12)+0.04VA+0.0125(VA-6) = 0.
VA(approx) = 4.3v.
0.02(VA-12)+0.04VA+0.0125(VA-6) = 0.
VA(approx) = 4.3v.
Sunilkumar said:
1 decade ago
@Shishir.
According to kcl, at node point, summation of all currents should be zero.
According to kcl, at node point, summation of all currents should be zero.
Kirthi said:
1 decade ago
For nodal analysis voltage source must be converted to current source right?
Shishir said:
1 decade ago
But surajit here you have solved taking the summation of current to be zero.
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