Malla Reddy College of Engineering and Technology (MRCET)
Department of EEE ( 2017-18 )
Electrical Circuits EEE
Solution:
The
node voltages and the directions of the branch currents are assigned as shown in given figure.
Applying KCL to node 1, we get:
5 = I10+ I3
5= (V1-0)/10 +(V1-V2)/3
V1(13/30) -V2(1/3) = 5 ..........(1)
Applying KCL to node 2, we get:
I3= I5 + I1
(V1-V2)/3 = (V2 -0)/5 + (V2-10) /1
V1(1/3)-V2(23/15) = -10................(2)
Solving the these two equations for V1 and V2 we get :
V1 = 19.85 V and V2 = 10.9 V and the currents are :
I10= V1/10 = 1.985A
I3 = (V1-V2)/3 = (19.85-10.9)/3 = 2.98A
I5 = V2/5 = 10.9/5 =2.18A
I1 = (V2-10) = (10.9-10)/1 = 0.9A
Super Node Analysis:
If there is only voltage source between two nodes in the given network
then it is difficult to apply the nodal analysis. Because the voltage source has to be converted into
a current source
in terms of the voltage source, write down the nodal
equations and relate the
node voltages to the voltage source. But this is a difficult approach .This
difficulty can be
avoided by creating super node which encloses the two nodes
that have common voltage
source.
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