Electrical Engineering - Branch, Loop and Node Analyses - Discussion
Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 10)
10.
Using the mesh current method, find the branch current, IR1, in the above figure.
Discussion:
9 comments Page 1 of 1.
Karthik Netha said:
2 years ago
Agree, thanks @Dipen.
Dipen said:
6 years ago
Using mesh law.
As we have two mesh,
from first mesh:
68(I1)+37(I1-I2)=12,
from 2nd mesh,
90(I2)+37(I2-I1)=-4,
Solving these two we get;
I1=0.1149Am.
I2=2mA.
As we have two mesh,
from first mesh:
68(I1)+37(I1-I2)=12,
from 2nd mesh,
90(I2)+37(I2-I1)=-4,
Solving these two we get;
I1=0.1149Am.
I2=2mA.
Prakashsb said:
6 years ago
How to solve this? Please tell me clearly.
Vijay said:
6 years ago
See they give mesh current but you solved nodal method how it's possible?
Subrotokumar said:
7 years ago
Thanks @Dilip.
Hassan said:
8 years ago
I solve the equation and the IR1 = 0.315 is that right?
Manju said:
8 years ago
Please explain it.
FAHAD said:
1 decade ago
Nodal equation.
For I1.
68I1+ (I1-I2) 37= 12V.
F0r I2.
90I2 + (I2-I1) 37= 4.
Solve these two equation simultaneously.
For I1.
68I1+ (I1-I2) 37= 12V.
F0r I2.
90I2 + (I2-I1) 37= 4.
Solve these two equation simultaneously.
Dilip said:
1 decade ago
Nodal eqn=>((va-12)/68)+(va/37)+((Va-4)/90)=0
Va=4.180volt
Ir1=((12-Va)/68)
= (12-4.18)/68
=115mA
Va=4.180volt
Ir1=((12-Va)/68)
= (12-4.18)/68
=115mA
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