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.
J.venkatesh said:
1 decade ago
Take rules according to kcl and kvl principles solve the problems.
Kirthi said:
1 decade ago
For nodal analysis voltage source must be converted to current source right?
Madhu said:
1 decade ago
I am not able to you understand. Can you explain me clearly? Please.
Ganesh said:
1 decade ago
@Krishna.
Is right according to nodal analysis. The algebraic sum of currents in a network of conductors meeting at a point is zero.
Here we take i=v/r.
Is right according to nodal analysis. The algebraic sum of currents in a network of conductors meeting at a point is zero.
Here we take i=v/r.
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).
Nagaraju N L said:
1 decade ago
I can't understand please solve the problem step by step please.
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.
Ram said:
10 years ago
Can you me proper sign convention in nodal analysis?
Ram said:
10 years ago
Sometimes V-12, sometimes 12-V what does it refer?
Abdullah U of R said:
10 years ago
Easy way to calculate node analysis:
i1+i2+i3 = 0 it is equal to:
(12-VA)/49 + -VA/24 + (6-VA)/80 = 0.
12/49 - VA/49 - VA/24 + 6/80 - VA/80 = 0.
12/49 + 6/80 = VA (1/49 + 1/24 + 1/80).
So, VA = (12/49 + 6/80)/(1/49 + 1/24 + 1/80) = 4.28 V.
i1+i2+i3 = 0 it is equal to:
(12-VA)/49 + -VA/24 + (6-VA)/80 = 0.
12/49 - VA/49 - VA/24 + 6/80 - VA/80 = 0.
12/49 + 6/80 = VA (1/49 + 1/24 + 1/80).
So, VA = (12/49 + 6/80)/(1/49 + 1/24 + 1/80) = 4.28 V.
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